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As a high school student myself, I would like to share some techniques:

$1.$ For integrals of the type

$$\int\frac{ae^x+be^{-x}}{pe^x+qe^{-x}}\mathrm dx$$

determine suitable constants $l$ and $m$ such that

$$pe^x+qe^{-x}=l(ae^x+be^{-x})+m(ae^x-be^{-x})$$

$2.$ (I find this particularly time-saving)For integrals involving $\ln x$ like

$$\int x\ln(\ln x)+\frac{x}{\ln^2x}\mathrm dx$$ $$\int\frac1{\ln x}-\frac1{\ln^2x}\mathrm dx$$

Substitute $\ln x=t\implies \mathrm dx=e^t\mathrm dt$, then apply the IBP formula that you stated to get the answer in a single step:

$$\int e^x\left(f(x)+f’(x)\right)\mathrm dx=e^xf(x)+C$$

$3.$ To evaluate $$\int(x-a)^m(x-b)^n\mathrm dx$$ where $m+n=-2$, take either of $(x-a)^2$ or $(x-b)^2$ common and substitute $\frac{x-b}{x-a}$ or $\frac{x-a}{x-b}$ respectively as $t$.

$4.$ To evaluate integrals of the following types, use $$\int P’(x)\mathrm dx=P(x)+C$$ as stated:

$$\begin{array}{c|cc}\text{Integrand}&{P(x)}\\\hline \frac1{(a+b\sin x)^2}&\frac{\cos x}{a+b\sin x}\\ \frac1{(a+b\cos x)^2}&\frac{\sin x}{a+b\cos x}\end{array}$$

$5.$ To evaluate integrals involving terms $(x\cos x+n\sin x)$ in the denominator of the integrand like

$$\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}\mathrm dx$$ $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}\mathrm dx$$

the compound angle formula proves to be very useful: $$x\cos x+n\sin x=\sqrt{x^2+n^2}\sin\left(x+\cot^{-1}\frac{x}{n}\right)$$

$6.$ Algebraic twins(as it’s known in my country):

To evaluate $$\int\frac{\mathrm dx}{x^4+kx^2+1}, \int\frac{x^2\mathrm dx}{x^4+kx^2+1}$$ write the numerator as $(x^2+1)\pm(x^2-1)$, then split into two integrals, divide by $x^2$ in the numerator and the denominator and recognize $\mathrm d\left(x\pm\frac1{x}\right)$.

This technique is also used to evaluate the trigonometric twins $$\int\sqrt{\tan x}\mathrm dx, \int\sqrt{\cot x}\mathrm dx $$

$7.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{\beta-x}$, substitute $x=\alpha\cos^2\theta+\beta\sin^2\theta$ or $x=\alpha\tanh^2\theta+\beta\text{sech}^2\theta$ or $x=\alpha\coth^2\theta+\beta\text{csch}^2\theta$.

$8.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{x-\beta}$ where $\beta>\alpha$, substitute $x=\beta\cosh^2\theta-\alpha\sinh^2\theta$ or $x=\alpha\tan^2\theta-\beta\sec^2\theta$ or $x=\alpha\cot^2\theta-\beta\csc^2\theta$ for $x\ge\beta$ and $x=\alpha\cosh^2\theta-\beta\sinh^2\theta$ or $x=\alpha\sec^2\theta-\beta\tan^2\theta$ or $x=\alpha\csc^2\theta-\beta\cot^2\theta$ for $x\le\alpha$.

$9.$ To evaluate integrals of the type $$\int e^{ax}\sin bx\mathrm dx, \int e^{ax}\cos bx\mathrm dx $$ $$\int_0^{2\pi}e^{\cos x}\cos(\sin x)\mathrm dx$$ use $$\cos\theta=\Re e^{\imath\theta}, \sin\theta=\Im e^{\imath\theta} $$ $10.$ $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ $11.$ Here is a list of general strategies to integrate irrational functions involving linear and quadratic polynomials:

$$\begin{array}{c|cc}\text{Integrand}&\text{Substitution}\\ \hline\int\frac{\mathrm dx}{L_1\sqrt{L_2}}&t=\sqrt L_2\\ \int\frac{\mathrm dx}{L\sqrt{Q}}&L=\frac1{t} \\ \int\frac{\mathrm dx}{Q\sqrt{L}}&t=\sqrt L\\ \int\frac{\mathrm dx}{Q_1\sqrt{Q_2}}&x=\frac1{t}\end{array}$$

$12.$ To evaluate integrals of the form $\int\sin^mx\cos^nx\mathrm dx$: $$\begin{array}{c|cc}\text{Condition}&\text{Technique}\\ \hline \text{m and n both are odd}&\text{Substitute the term with the higher power as }t.\\ \text{One is odd, the other even.}&\text{Substitute the term with the odd power as }t.\\ \text{m and n both are even}&\text{Reduce the powers by using multiple angle formulae or De-Moivre’s theorem.}\\ m,n\in\mathbb Q^-\text{ such that }m+n\in\mathbb Z^-& \text{Take }\sec^{-m-n}x\text{ common and substitute }t=\tan x.\end{array}$$

As a high school student myself, I would like to share some techniques:

$1.$ For integrals of the type

$$\int\frac{ae^x+be^{-x}}{pe^x+qe^{-x}}\mathrm dx$$

determine suitable constants $l$ and $m$ such that

$$pe^x+qe^{-x}=l(ae^x+be^{-x})+m(ae^x-be^{-x})$$

$2.$ (I find this particularly time-saving)For integrals involving $\ln x$ like

$$\int x\ln(\ln x)+\frac{x}{\ln^2x}\mathrm dx$$ $$\int\frac1{\ln x}-\frac1{\ln^2x}\mathrm dx$$

Substitute $\ln x=t\implies \mathrm dx=e^t\mathrm dt$, then apply the IBP formula that you stated to get the answer in a single step:

$$\int e^x\left(f(x)+f’(x)\right)\mathrm dx=e^xf(x)+C$$

$3.$ To evaluate $$\int(x-a)^m(x-b)^n\mathrm dx$$ where $m+n=-2$, take either of $(x-a)^2$ or $(x-b)^2$ common and substitute $\frac{x-b}{x-a}$ or $\frac{x-a}{x-b}$ respectively as $t$.

$4.$ To evaluate integrals of the following types, use $$\int P’(x)\mathrm dx=P(x)+C$$ as stated:

$$\begin{array}{c|cc}\text{Integrand}&{P(x)}\\\hline \frac1{(a+b\sin x)^2}&\frac{\cos x}{a+b\sin x}\\ \frac1{(a+b\cos x)^2}&\frac{\sin x}{a+b\cos x}\end{array}$$

$5.$ To evaluate integrals involving terms $(x\cos x+n\sin x)$ in the denominator of the integrand like

$$\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}\mathrm dx$$ $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}\mathrm dx$$

the compound angle formula proves to be very useful: $$x\cos x+n\sin x=\sqrt{x^2+n^2}\sin\left(x+\cot^{-1}\frac{x}{n}\right)$$

$6.$ Algebraic twins(as it’s known in my country):

To evaluate $$\int\frac{\mathrm dx}{x^4+kx^2+1}, \int\frac{x^2\mathrm dx}{x^4+kx^2+1}$$ write the numerator as $(x^2+1)\pm(x^2-1)$, then split into two integrals, divide by $x^2$ in the numerator and the denominator and recognize $\mathrm d\left(x\pm\frac1{x}\right)$.

This technique is also used to evaluate the trigonometric twins $$\int\sqrt{\tan x}\mathrm dx, \int\sqrt{\cot x}\mathrm dx $$

$7.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{\beta-x}$, substitute $x=\alpha\cos^2\theta+\beta\sin^2\theta$ or $x=\alpha\tanh^2\theta+\beta\text{sech}^2\theta$ or $x=\alpha\coth^2\theta+\beta\text{csch}^2\theta$.

$8.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{x-\beta}$ where $\beta>\alpha$, substitute $x=\beta\cosh^2\theta-\alpha\sinh^2\theta$ or $x=\alpha\tan^2\theta-\beta\sec^2\theta$ for $x\ge\beta$ and $x=\alpha\cosh^2\theta-\beta\sinh^2\theta$ or $x=\alpha\sec^2\theta-\beta\tan^2\theta$ for $x\le\alpha$.

$9.$ To evaluate integrals of the type $$\int e^{ax}\sin bx\mathrm dx, \int e^{ax}\cos bx\mathrm dx $$ $$\int_0^{2\pi}e^{\cos x}\cos(\sin x)\mathrm dx$$ use $$\cos\theta=\Re e^{\imath\theta}, \sin\theta=\Im e^{\imath\theta} $$ $10.$ $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ $11.$ Here is a list of general strategies to integrate irrational functions involving linear and quadratic polynomials:

$$\begin{array}{c|cc}\text{Integrand}&\text{Substitution}\\ \hline\int\frac{\mathrm dx}{L_1\sqrt{L_2}}&t=\sqrt L_2\\ \int\frac{\mathrm dx}{L\sqrt{Q}}&L=\frac1{t} \\ \int\frac{\mathrm dx}{Q\sqrt{L}}&t=\sqrt L\\ \int\frac{\mathrm dx}{Q_1\sqrt{Q_2}}&x=\frac1{t}\end{array}$$

$12.$ To evaluate integrals of the form $\int\sin^mx\cos^nx\mathrm dx$: $$\begin{array}{c|cc}\text{Condition}&\text{Technique}\\ \hline \text{m and n both are odd}&\text{Substitute the term with the higher power as }t.\\ \text{One is odd, the other even.}&\text{Substitute the term with the odd power as }t.\\ \text{m and n both are even}&\text{Reduce the powers by using multiple angle formulae or De-Moivre’s theorem.}\\ m,n\in\mathbb Q^-\text{ such that }m+n\in\mathbb Z^-& \text{Take }\sec^{-m-n}x\text{ common and substitute }t=\tan x.\end{array}$$

As a high school student myself, I would like to share some techniques:

$1.$ For integrals of the type

$$\int\frac{ae^x+be^{-x}}{pe^x+qe^{-x}}\mathrm dx$$

determine suitable constants $l$ and $m$ such that

$$pe^x+qe^{-x}=l(ae^x+be^{-x})+m(ae^x-be^{-x})$$

$2.$ (I find this particularly time-saving)For integrals involving $\ln x$ like

$$\int x\ln(\ln x)+\frac{x}{\ln^2x}\mathrm dx$$ $$\int\frac1{\ln x}-\frac1{\ln^2x}\mathrm dx$$

Substitute $\ln x=t\implies \mathrm dx=e^t\mathrm dt$, then apply the IBP formula that you stated to get the answer in a single step:

$$\int e^x\left(f(x)+f’(x)\right)\mathrm dx=e^xf(x)+C$$

$3.$ To evaluate $$\int(x-a)^m(x-b)^n\mathrm dx$$ where $m+n=-2$, take either of $(x-a)^2$ or $(x-b)^2$ common and substitute $\frac{x-b}{x-a}$ or $\frac{x-a}{x-b}$ respectively as $t$.

$4.$ To evaluate integrals of the following types, use $$\int P’(x)\mathrm dx=P(x)+C$$ as stated:

$$\begin{array}{c|cc}\text{Integrand}&{P(x)}\\\hline \frac1{(a+b\sin x)^2}&\frac{\cos x}{a+b\sin x}\\ \frac1{(a+b\cos x)^2}&\frac{\sin x}{a+b\cos x}\end{array}$$

$5.$ To evaluate integrals involving terms $(x\cos x+n\sin x)$ in the denominator of the integrand like

$$\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}\mathrm dx$$ $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}\mathrm dx$$

the compound angle formula proves to be very useful: $$x\cos x+n\sin x=\sqrt{x^2+n^2}\sin\left(x+\cot^{-1}\frac{x}{n}\right)$$

$6.$ Algebraic twins(as it’s known in my country):

To evaluate $$\int\frac{\mathrm dx}{x^4+kx^2+1}, \int\frac{x^2\mathrm dx}{x^4+kx^2+1}$$ write the numerator as $(x^2+1)\pm(x^2-1)$, then split into two integrals, divide by $x^2$ in the numerator and the denominator and recognize $\mathrm d\left(x\pm\frac1{x}\right)$.

This technique is also used to evaluate the trigonometric twins $$\int\sqrt{\tan x}\mathrm dx, \int\sqrt{\cot x}\mathrm dx $$

$7.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{\beta-x}$, substitute $x=\alpha\cos^2\theta+\beta\sin^2\theta$.

$8.$ To evaluate integrals of the type $$\int e^{ax}\sin bx\mathrm dx, \int e^{ax}\cos bx\mathrm dx $$ $$\int_0^{2\pi}e^{\cos x}\cos(\sin x)\mathrm dx$$ use $$\cos\theta=\Re e^{\imath\theta}, \sin\theta=\Im e^{\imath\theta} $$ $9.$ $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ $10.$ Here is a list of general strategies to integrate irrational functions involving linear and quadratic polynomials:

$$\begin{array}{c|cc}\text{Integrand}&\text{Substitution}\\ \hline\int\frac{\mathrm dx}{L_1\sqrt{L_2}}&t=\sqrt L_2\\ \int\frac{\mathrm dx}{L\sqrt{Q}}&L=\frac1{t} \\ \int\frac{\mathrm dx}{Q\sqrt{L}}&t=\sqrt L\\ \int\frac{\mathrm dx}{Q_1\sqrt{Q_2}}&x=\frac1{t}\end{array}$$

$11.$ To evaluate integrals of the form $\int\sin^mx\cos^nx\mathrm dx$: $$\begin{array}{c|cc}\text{Condition}&\text{Technique}\\ \hline \text{m and n both are odd}&\text{Substitute the term with the higher power as }t.\\ \text{One is odd, the other even.}&\text{Substitute the term with the odd power as }t.\\ \text{m and n both are even}&\text{Reduce the powers by using multiple angle formulae or De-Moivre’s theorem.}\\ m,n\in\mathbb Q^-\text{ such that }m+n\in\mathbb Z^-& \text{Take }\sec^{-m-n}x\text{ common and substitute }t=\tan x.\end{array}$$

As a high school student myself, I would like to share some techniques:

$1.$ For integrals of the type

$$\int\frac{ae^x+be^{-x}}{pe^x+qe^{-x}}\mathrm dx$$

determine suitable constants $l$ and $m$ such that

$$pe^x+qe^{-x}=l(ae^x+be^{-x})+m(ae^x-be^{-x})$$

$2.$ (I find this particularly time-saving)For integrals involving $\ln x$ like

$$\int x\ln(\ln x)+\frac{x}{\ln^2x}\mathrm dx$$ $$\int\frac1{\ln x}-\frac1{\ln^2x}\mathrm dx$$

Substitute $\ln x=t\implies \mathrm dx=e^t\mathrm dt$, then apply the IBP formula that you stated to get the answer in a single step:

$$\int e^x\left(f(x)+f’(x)\right)\mathrm dx=e^xf(x)+C$$

$3.$ To evaluate $$\int(x-a)^m(x-b)^n\mathrm dx$$ where $m+n=-2$, take either of $(x-a)^2$ or $(x-b)^2$ common and substitute $\frac{x-b}{x-a}$ or $\frac{x-a}{x-b}$ respectively as $t$.

$4.$ To evaluate integrals of the following types, use $$\int P’(x)\mathrm dx=P(x)+C$$ as stated:

$$\begin{array}{c|cc}\text{Integrand}&{P(x)}\\\hline \frac1{(a+b\sin x)^2}&\frac{\cos x}{a+b\sin x}\\ \frac1{(a+b\cos x)^2}&\frac{\sin x}{a+b\cos x}\end{array}$$

$5.$ To evaluate integrals involving terms $(x\cos x+n\sin x)$ in the denominator of the integrand like

$$\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}\mathrm dx$$ $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}\mathrm dx$$

the compound angle formula proves to be very useful: $$x\cos x+n\sin x=\sqrt{x^2+n^2}\sin\left(x+\cot^{-1}\frac{x}{n}\right)$$

$6.$ Algebraic twins(as it’s known in my country):

To evaluate $$\int\frac{\mathrm dx}{x^4+kx^2+1}, \int\frac{x^2\mathrm dx}{x^4+kx^2+1}$$ write the numerator as $(x^2+1)\pm(x^2-1)$, then split into two integrals, divide by $x^2$ in the numerator and the denominator and recognize $\mathrm d\left(x\pm\frac1{x}\right)$.

This technique is also used to evaluate the trigonometric twins $$\int\sqrt{\tan x}\mathrm dx, \int\sqrt{\cot x}\mathrm dx $$

$7.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{\beta-x}$, substitute $x=\alpha\cos^2\theta+\beta\sin^2\theta$ or $x=\alpha\tanh^2\theta+\beta\text{sech}^2\theta$ or $x=\alpha\coth^2\theta+\beta\text{csch}^2\theta$.

$8.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{x-\beta}$ where $\beta>\alpha$, substitute $x=\beta\cosh^2\theta-\alpha\sinh^2\theta$ or $x=\alpha\tan^2\theta-\beta\sec^2\theta$ or $x=\alpha\cot^2\theta-\beta\csc^2\theta$ for $x\ge\beta$ and $x=\alpha\cosh^2\theta-\beta\sinh^2\theta$ or $x=\alpha\sec^2\theta-\beta\tan^2\theta$ or $x=\alpha\csc^2\theta-\beta\cot^2\theta$ for $x\le\alpha$.

$9.$ To evaluate integrals of the type $$\int e^{ax}\sin bx\mathrm dx, \int e^{ax}\cos bx\mathrm dx $$ $$\int_0^{2\pi}e^{\cos x}\cos(\sin x)\mathrm dx$$ use $$\cos\theta=\Re e^{\imath\theta}, \sin\theta=\Im e^{\imath\theta} $$ $10.$ $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ $11.$ Here is a list of general strategies to integrate irrational functions involving linear and quadratic polynomials:

$$\begin{array}{c|cc}\text{Integrand}&\text{Substitution}\\ \hline\int\frac{\mathrm dx}{L_1\sqrt{L_2}}&t=\sqrt L_2\\ \int\frac{\mathrm dx}{L\sqrt{Q}}&L=\frac1{t} \\ \int\frac{\mathrm dx}{Q\sqrt{L}}&t=\sqrt L\\ \int\frac{\mathrm dx}{Q_1\sqrt{Q_2}}&x=\frac1{t}\end{array}$$

$12.$ To evaluate integrals of the form $\int\sin^mx\cos^nx\mathrm dx$: $$\begin{array}{c|cc}\text{Condition}&\text{Technique}\\ \hline \text{m and n both are odd}&\text{Substitute the term with the higher power as }t.\\ \text{One is odd, the other even.}&\text{Substitute the term with the odd power as }t.\\ \text{m and n both are even}&\text{Reduce the powers by using multiple angle formulae or De-Moivre’s theorem.}\\ m,n\in\mathbb Q^-\text{ such that }m+n\in\mathbb Z^-& \text{Take }\sec^{-m-n}x\text{ common and substitute }t=\tan x.\end{array}$$

As a high school student myself, I would like to share some techniques:

$1.$ For integrals of the type

$$\int\frac{ae^x+be^{-x}}{pe^x+qe^{-x}}\mathrm dx$$

determine suitable constants $l$ and $m$ such that

$$pe^x+qe^{-x}=l(ae^x+be^{-x})+m(ae^x-be^{-x})$$

$2.$ (I find this particularly time-saving)For integrals involving $\ln x$ like

$$\int x\ln(\ln x)+\frac{x}{\ln^2x}\mathrm dx$$ $$\int\frac1{\ln x}-\frac1{\ln^2x}\mathrm dx$$

Substitute $\ln x=t\implies \mathrm dx=e^t\mathrm dt$, then apply the IBP formula that you stated to get the answer in a single step:

$$\int e^x\left(f(x)+f’(x)\right)\mathrm dx=e^xf(x)+C$$

$3.$ To evaluate $$\int(x-a)^m(x-b)^n\mathrm dx$$ where $m+n=-2$, take either of $(x-a)^2$ or $(x-b)^2$ common and substitute $\frac{x-b}{x-a}$ or $\frac{x-a}{x-b}$ respectively as $t$.

$4.$ To evaluate integrals of the following types, use $$\int P’(x)\mathrm dx=P(x)+C$$ as stated:

$$\begin{array}{c|cc}\text{Integrand}&{P(x)}\\\hline \frac1{(a+b\sin x)^2}&\frac{\cos x}{a+b\sin x}\\ \frac1{(a+b\cos x)^2}&\frac{\sin x}{a+b\cos x}\end{array}$$

$5.$ To evaluate integrals involving terms $(x\cos x+n\sin x)$ in the denominator of the integrand like

$$\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}\mathrm dx$$ $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}\mathrm dx$$

the compound angle formula proves to be very useful: $$x\cos x+n\sin x=\sqrt{x^2+n^2}\sin\left(x+\cot^{-1}\frac{x}{n}\right)$$

$6.$ Algebraic twins(as it’s known in my country):

To evaluate $$\int\frac{\mathrm dx}{x^4+kx^2+1}, \int\frac{x^2\mathrm dx}{x^4+kx^2+1}$$ write the numerator as $(x^2+1)\pm(x^2-1)$, then split into two integrals, divide by $x^2$ in the numerator and the denominator and recognize $\mathrm d\left(x\pm\frac1{x}\right)$.

This technique is also used to evaluate the trigonometric twins $$\int\sqrt{\tan x}\mathrm dx, \int\sqrt{\cot x}\mathrm dx $$

$7.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{\beta-x}$, substitute $x=\alpha\cos^2\theta+\beta\sin^2\theta$.

$8.$ To evaluate integrals of the type $$\int e^{ax}\sin bx\mathrm dx, \int e^{ax}\cos bx\mathrm dx $$ $$\int_0^{2\pi}e^{\cos x}\cos(\sin x)\mathrm dx$$ use $$\cos\theta=\Re e^{\imath\theta}, \sin\theta=\Im e^{\imath\theta} $$ $9.$ $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$

As a high school student myself, I would like to share some techniques:

$1.$ For integrals of the type

$$\int\frac{ae^x+be^{-x}}{pe^x+qe^{-x}}\mathrm dx$$

determine suitable constants $l$ and $m$ such that

$$pe^x+qe^{-x}=l(ae^x+be^{-x})+m(ae^x-be^{-x})$$

$2.$ (I find this particularly time-saving)For integrals involving $\ln x$ like

$$\int x\ln(\ln x)+\frac{x}{\ln^2x}\mathrm dx$$ $$\int\frac1{\ln x}-\frac1{\ln^2x}\mathrm dx$$

Substitute $\ln x=t\implies \mathrm dx=e^t\mathrm dt$, then apply the IBP formula that you stated to get the answer in a single step:

$$\int e^x\left(f(x)+f’(x)\right)\mathrm dx=e^xf(x)+C$$

$3.$ To evaluate $$\int(x-a)^m(x-b)^n\mathrm dx$$ where $m+n=-2$, take either of $(x-a)^2$ or $(x-b)^2$ common and substitute $\frac{x-b}{x-a}$ or $\frac{x-a}{x-b}$ respectively as $t$.

$4.$ To evaluate integrals of the following types, use $$\int P’(x)\mathrm dx=P(x)+C$$ as stated:

$$\begin{array}{c|cc}\text{Integrand}&{P(x)}\\\hline \frac1{(a+b\sin x)^2}&\frac{\cos x}{a+b\sin x}\\ \frac1{(a+b\cos x)^2}&\frac{\sin x}{a+b\cos x}\end{array}$$

$5.$ To evaluate integrals involving terms $(x\cos x+n\sin x)$ in the denominator of the integrand like

$$\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}\mathrm dx$$ $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}\mathrm dx$$

the compound angle formula proves to be very useful: $$x\cos x+n\sin x=\sqrt{x^2+n^2}\sin\left(x+\cot^{-1}\frac{x}{n}\right)$$

$6.$ Algebraic twins(as it’s known in my country):

To evaluate $$\int\frac{\mathrm dx}{x^4+kx^2+1}, \int\frac{x^2\mathrm dx}{x^4+kx^2+1}$$ write the numerator as $(x^2+1)\pm(x^2-1)$, then split into two integrals, divide by $x^2$ in the numerator and the denominator and recognize $\mathrm d\left(x\pm\frac1{x}\right)$.

This technique is also used to evaluate the trigonometric twins $$\int\sqrt{\tan x}\mathrm dx, \int\sqrt{\cot x}\mathrm dx $$

$7.$ For evaluating integrals that involve $\sqrt{x-\alpha}$ and $\sqrt{\beta-x}$, substitute $x=\alpha\cos^2\theta+\beta\sin^2\theta$.

$8.$ To evaluate integrals of the type $$\int e^{ax}\sin bx\mathrm dx, \int e^{ax}\cos bx\mathrm dx $$ $$\int_0^{2\pi}e^{\cos x}\cos(\sin x)\mathrm dx$$ use $$\cos\theta=\Re e^{\imath\theta}, \sin\theta=\Im e^{\imath\theta} $$ $9.$ $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ $10.$ Here is a list of general strategies to integrate irrational functions involving linear and quadratic polynomials:

$$\begin{array}{c|cc}\text{Integrand}&\text{Substitution}\\ \hline\int\frac{\mathrm dx}{L_1\sqrt{L_2}}&t=\sqrt L_2\\ \int\frac{\mathrm dx}{L\sqrt{Q}}&L=\frac1{t} \\ \int\frac{\mathrm dx}{Q\sqrt{L}}&t=\sqrt L\\ \int\frac{\mathrm dx}{Q_1\sqrt{Q_2}}&x=\frac1{t}\end{array}$$

$11.$ To evaluate integrals of the form $\int\sin^mx\cos^nx\mathrm dx$: $$\begin{array}{c|cc}\text{Condition}&\text{Technique}\\ \hline \text{m and n both are odd}&\text{Substitute the term with the higher power as }t.\\ \text{One is odd, the other even.}&\text{Substitute the term with the odd power as }t.\\ \text{m and n both are even}&\text{Reduce the powers by using multiple angle formulae or De-Moivre’s theorem.}\\ m,n\in\mathbb Q^-\text{ such that }m+n\in\mathbb Z^-& \text{Take }\sec^{-m-n}x\text{ common and substitute }t=\tan x.\end{array}$$