# Antiderivative (Indefinite Integral)

Consider a continuous function $ƒ$ defined in an open interval $(a,b)$. Then a function $F$ is said to be an antiderivative of $ƒ$ on the interval, if the derivative of $F$ is equal to $ƒ$ on the interval i.e. $\text{if }\frac{dF(x)}{dx}=ƒ(x),x\in(a,b)$

As the derivative of a constant is zero, $F(x)+C$ is also an antiderivative of $ƒ$. Thus, when $\frac{dF(x)}{dx}=ƒ(x)$, we have $\frac{d[F(x)+C]}{dx}=\frac{dF(x)}{dx}+\frac{dC}{dx}$ $=ƒ(x)+0$ $=ƒ(x)$

Any two antiderivatives of a function also differ by a constant. Let $F$ and $G$ be the antiderivatives of a function $ƒ$. Then, $\frac{d[F(x)-G(x)]}{dx}=\frac{dF(x)}{dx}-\frac{dG(x)}{dx}$ $=ƒ(x)-ƒ(x)$ $=0$ Hence, from this, it follows that there exists a constant $C$ such that $F(x)-G(x)=C$

Hence, all the above mentioned things establish the fact that if $F$ is an antiderivative of $ƒ$, $F(x)+C$ gives all the possible antiderivatives of $ƒ$, when $C$ runs through all real numbers.

The general form of all antiderivatives of $ƒ$, which we call indefinite integral of $ƒ$, is denoted by $\int ƒdx \text{ or } \int ƒ(x)dx$ If $F$ is an antiderivative of $ƒ$, we have $\int ƒ(x)dx=F(x)+C$

One basic property of the indefinite integral is $\int [c_1ƒ(x)+c_2g(x)]dx$$=c_1\int ƒ(x)dx+c_2\int g(x)dx+C$ where $ƒ$ and $g$ are continuous functions in an interval $(a,b)$ and $c_1$ and $c_2$ are some constants.

## Some Integration Formulae

$\int 1dx=x+C$ $\int x^ndx=\frac{x^{n+1}}{n+1}+C\text{ (n≠-1)}$ $\int\sin axdx=-\frac{\cos ax}{a}+C$ $\int\cos axdx=\frac{\sin ax}{a}+C$ $\int\sec^2axdx=\frac{\tan ax}{a}+C$ $\int\operatorname{cosec}^2axdx=-\frac{\cot ax}{a}+C$ $\int\sec ax\tan ax dx=\frac{\sec ax}{a}+C$ $\int\operatorname{cosec}ax\cot axdx=-\frac{\operatorname{cosec}ax}{a}+C$ $\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{(n+1)a}+C$ $\int\frac{1}{x}dx=\log x+C$ $\int e^{ax}dx=\frac{e^{ax}}{a}+C$ $\int\frac{1}{ax+b}dx=\frac{\log(ax+b)}{a}+C$

## Find the indefinite integrals:

### $\int 5x^3dx$

$\int 5x^3dx$ $=5\int x^3dx$ $=\frac{5}{4}x^4+C$

### $\int (2x+1)(3x+2)dx$

$\int (2x+1)(3x+2)dx$ $=\int (6x^2+7x+2)dx$ $=\int 6x^2dx+\int 7xdx+\int 2 dx$ $=6\frac{x^3}{3}+7\frac{x^2}{2}+2x+C$ $=2x^3+\frac{7}{2}x^2+2x+C$

### $\int\frac{ax^2+bx+c}{x^2}dx$

$\int\frac{ax^2+bx+c}{x^2}dx$ $=\left(a+\frac{b}{x}+\frac{c}{x^2}\right)dx$ $=\int adx+\int\frac{b}{x}dx+\frac{c}{x^2}dx$ $=a\int 1dx+b\int\frac{1}{x}dx+c\int\frac{1}{x^2}dx$ $=ax+b\log x+c\frac{x^{-1}}{-1}+C$ $=ax+b\log x-\frac{c}{x}+C$

### $\int(a-bx)^5dx$

$\int(a-bx)^5dx$ $=\frac{(a-bx)^6}{6×(-b)}+C$ $=-\frac{1}{6b}(a-bx)^6+C$

### $\int\frac{x^2+5}{x+2}dx$

$\int\frac{x^5+5}{x+2}dx$ $=\int\frac{\overline{x^2-4}+9}{x+2}dx$ $=\int\frac{(x+2)(x-2)+9}{x+2}dx$ $=\int(x-2)dx+\int\frac{9}{x+2}dx$ $=\int xdx-\int 2dx+9\frac{1}{x+2}dx$ $=\frac{x^2}{2}-2x+9\log(x+2)+C$

### $\int 2x\sqrt{2x+3}dx$

$\int 2x\sqrt{2x+3}dx$ $=\int (\overline{2x+3}-3)\sqrt{2x+3}dx$ $=\int(2x+3)^{\frac{3}{2}}dx-\int 3(2x+3)^{\frac{1}{2}}dx$ $=\frac{(2x+3)^{\frac{5}{2}}}{\frac{5}{2}×2}-3\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}×2}+C$ $=\frac{1}{5}(2x+3)^{\frac{5}{2}}-(2x+3)^{\frac{3}{2}}+C$

### $\int\frac{3x+2}{\sqrt{5x+3}}dx$

$\int\frac{3x+2}{\sqrt{5x+3}}dx$ $=3\int\frac{x+\frac{2}{3}}{\sqrt{5x+3}}dx$ $=\frac{3}{5}\int\frac{5x+\frac{10}{3}}{\sqrt{5x+3}}dx$ $=\frac{3}{5}\int\frac{\overline{5x+3}+\frac{1}{3}}{\sqrt{5x+3}}dx$ $=\frac{3}{5}\int(5x+3)^{\frac{1}{2}}dx+\frac{3}{5}\int\frac{1}{3}(5x+3)^{-\frac{1}{2}}dx$ $=\frac{3}{5}\frac{(5x+3)^{\frac{3}{2}}}{\frac{3}{2}×5}+\frac{1}{5}\frac{(5x+3)^{\frac{1}{2}}}{\frac{1}{2}×5}+C$ $=\frac{2}{25}(5x+3)^{\frac{3}{2}}+\frac{2}{25}(5x+3)^{\frac{1}{2}}+C$

### $\int\frac{dx}{\sqrt{x+a}-\sqrt{x-b}}$

$\int\frac{dx}{\sqrt{x+a}-\sqrt{x-b}}$ $=\int\frac{\sqrt{x+a}+\sqrt{x-b}}{x+a-x+b}dx$ $=\int\frac{\sqrt{x+a}+\sqrt{x-b}}{a+b}dx$$=\frac{1}{a+b}\left[\int(x+a)^{\frac{1}{2}}dx+\int(x-b)^{\frac{1}{2}}dx\right]$ $=\frac{1}{a+b}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x-b)^{\frac{3}{2}}}{\frac{3}{2}}\right]+C$ $=\frac{2}{3(a+b)}\left[(x+a)^{\frac{3}{2}}+(x-b)^{\frac{3}{2}}\right]+C$

### $\int\cos^2bxdx$

$\int\cos^2bxdx$ $=\frac{1}{2}\int 2\cos^2bxdx$ $=\frac{1}{2}\int(1+\cos 2bx)dx$ $=\frac{1}{2}\int 1dx+\frac{1}{2}\int\cos 2bx dx$ $=\frac{1}{2}x+\frac{1}{2}\frac{\sin 2bx}{2b}+C$ $=\frac{x}{2}+\frac{1}{4b}\sin 2bx+C$

### $\int\frac{1}{\cos^2x\sin^2x}dx$

$\int\frac{1}{\cos^2x\sin^2x}dx$ $=\int\frac{\sin^2x+\cos^2x}{\cos^2x\sin^2x}dx$ $=\int\sec^2xdx+\int\operatorname{cosec}^2xdx$ $=\tan x-\cot x+C$

### $\int\sqrt{1+\cos nx}dx$

$\int\sqrt{1+\cos nx}dx$ $=\int\sqrt{2\cos^2\frac{nx}{2}}dx$ $=\sqrt{2}\int\cos\frac{nx}{2}dx$ $=\sqrt{2}\frac{\sin\frac{nx}{2}}{\frac{n}{2}}+C$ $=\frac{2\sqrt{2}}{n}\sin\frac{nx}{2}+C$

### $\int\frac{dx}{1-\sin ax}$

$\int\frac{dx}{1-\sin ax}$ $=\int\frac{1+\sin ax}{\cos^2ax}dx$ $=\int\sec^2axdx+\int\sec ax\tan ax dx$ $=\frac{\tan ax}{a}+\frac{\sec ax}{a}+C$

### $\int\cos px\cos qxdx$

$\int\cos px\cos qxdx$ $=\frac{1}{2}\int 2\cos px\cos qxdx$ $=\frac{1}{2}\int[\cos(px-qx)+\cos(px+qx)]dx$ $=\frac{1}{2}\int[\cos(p-q)x+\cos(p+q)x]dx$ $=-\frac{1}{2}\frac{\sin(p-q)x}{p-q}-\frac{1}{2}\frac{\sin(p+q)x}{p+q}+C$ $=-\frac{\sin(p-q)x}{2(p-q)}-\frac{\sin(p+q)x}{2(p+q)}+C$

### $\int(e^{px}+e^{-px})^2dx$

$\int(e^{px}+e^{-px})^2dx$ $=\int(e^{2px}+2+e^{-2px})dx$ $=\int e^{2px}dx+2\int 1dx+\int e^{-2px}dx$ $=\frac{e^{2px}}{2p}+2x+\frac{e^{-2px}}{-2p}+C$ $=\frac{1}{2p}e^{2px}+2x-\frac{1}{2p}e^{-2px}+C$

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