Indefinite Integration - Notes, Concept and All Important Formula

INDEFINITE INTEGRATION

If  f & F are function of $x$ such that $F^{\prime}(x)$$=f(x)$ then the function $F$ is called a PRIMITIVE OR ANTIDERIVATIVE OR INTEGRAL of $\mathrm{f}(\mathrm{x})$ w.r.t. $\mathrm{x}$ and is written symbolically as $\displaystyle \int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$$=\mathrm{F}(\mathrm{x})+\mathrm{c} \Leftrightarrow \dfrac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{F}(\mathrm{x})+\mathrm{c}\}$$=\mathrm{f}(\mathrm{x})$, where $\mathrm{c}$ is called the constant of integration.

Note : If $\displaystyle \int f(x) d x$$=F(x)+c$, then $\displaystyle \int f(a x+b) d x$$=\dfrac{F(a x+b)}{a}+c, a \neq 0$

---

All Chapter Notes, Concept and Important Formula

---

1. STANDARD RESULTS :

(i) $\displaystyle \int(a x+b)^{n} d x$$=\dfrac{(a x+b)^{n+1}}{a(n+1)}+c ; n \neq-1$

(ii) $\displaystyle \int \dfrac{d x}{a x+b}$$=\dfrac{1}{a} \ln|a x+b|+c$

(iii) $\displaystyle \int e^{\mathrm{ax}+b} \mathrm{dx}$$=\dfrac{1}{\mathrm{a}} e^{\mathrm{ax}+\mathrm{b}}+\mathrm{c}$

(iv) $\displaystyle \int a^{p x+q} d x$$=\dfrac{1}{p} \dfrac{a^{p x+q}}{\ln a}(a>0)+c$

(v) $\displaystyle \int \sin (a x+b) d x$$=-\dfrac{1}{a} \cos (a x+b)+c$

(vi) $\displaystyle \int \cos (a x+b) d x$$=\dfrac{1}{a} \sin (a x+b)+c$

(vii) $\displaystyle \int \tan (a x+b) d x$$=\dfrac{1}{a} \ln|\sec (a x+b)|+c$

(viii) $\displaystyle \int \cot (a x+b) d x$$=\dfrac{1}{a} \ln|\sin (a x+b)|+c$

(ix) $\displaystyle \int \sec ^{2}(a x+b) d x$$=\dfrac{1}{a} \tan (a x+b)+c$

(x) $\displaystyle \int \operatorname{cosec}^{2}(a x+b) d x$$=-\dfrac{1}{a} \cot (a x+b)+c$

(xi) $\displaystyle \int \operatorname{cosec}(a x+b) \cdot \cot (a x+b) d x$$=-\dfrac{1}{a} \operatorname{cosec}(a x+b)+c$

(xii) $\displaystyle \int \sec (a x+b) \cdot \tan (a x+b) d x$$=\dfrac{1}{a} \cdot \sec (a x+b)+c$

(xiii) $\displaystyle \int \sec x d x$$=\ln |\sec x+\tan x|+c$

OR $\displaystyle \int \sec x d x$$=\ln \tan \left|\dfrac{\pi}{4}+\dfrac{x}{2}\right|+c$

(xiv) $\displaystyle \int \operatorname{cosec} x d x$$=\ln|\operatorname{cosec} x-\cot x|+c$

OR $\displaystyle \int \operatorname{cosec} x d x$$=\ln \left|\tan \dfrac{x}{2}\right|+c$ OR $-\ln(\operatorname{cosec} x+\cot x)+c$

(xv) $\displaystyle \int \dfrac{d x}{\sqrt{a^{2}-x^{2}}}$$=\sin ^{-1} \dfrac{x}{a}+c$

(xvi) $\displaystyle \int \dfrac{d x}{a^{2}+x^{2}}$$=\dfrac{1}{a} \tan ^{-1} \dfrac{x}{a}+c$

(xvii) $\displaystyle \int \dfrac{d x}{x \sqrt{x^{2}-a^{2}}}$$=\dfrac{1}{a} \sec ^{-1} \dfrac{x}{a}+c$

(xviii) $\displaystyle \int \dfrac{d x}{\sqrt{x^{2}+a^{2}}}$$=\ln\left[x+\sqrt{x^{2}+a^{2}}\right]+c$

(xix) $\displaystyle \int \dfrac{d x}{\sqrt{x^{2}-a^{2}}}$$=\ln\left[x+\sqrt{x^{2}-a^{2}}\right]+c$

(xx) $\displaystyle \int \dfrac{d x}{a^{2}-x^{2}}$$=\dfrac{1}{2 a} \ln \left|\dfrac{a+x}{a-x}\right|+c$

(xxi) $\displaystyle \int \dfrac{d x}{x^{2}-a^{2}}$$=\dfrac{1}{2 a} \ln \left|\dfrac{x-a}{x+a}\right|+c$

(xxii) $\displaystyle \int \sqrt{a^{2}-x^{2}} d x$$=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}+c$

(xxiii) $\displaystyle \int \sqrt{x^{2}+a^{2}} d x$$=\dfrac{x}{2} \sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}+a^{2}}\right)+c$

(xxiv) $\displaystyle \int \sqrt{x^{2}-a^{2}} d x$$=\dfrac{x}{2} \sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}-a^{2}}\right)+c$

(xxv) $\displaystyle \int e^{\mathrm{ax}} \cdot \sin b x d x$$=\dfrac{e^{\operatorname{ax}}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+c$

(xxvi) $\displaystyle \int e^{a x} \cdot \cos b x d x$$=\dfrac{e^{\operatorname{ax}}}{a^{2}+b^{2}}(a \cos b x+b \sin b x)+c$

---

---

2. TECHNIQUES OF INTEGRATION :

(a) Substitution or change of independent variable:

Integral $\mathrm{I}$$=\displaystyle \int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$ is changed to $\displaystyle \int \mathrm{f}(\phi(\mathrm{t})) \phi^{\prime}(\mathrm{t})\,\,  \mathrm{dt}$, by a suitable substitution $\mathrm{x}$$=\phi$ (t) provided the later integral is easier to integrate.

Some standard substitution:

(1) $\displaystyle \int[f(x)]^{n} f^{\prime}(x) d x \quad O R \displaystyle \int \dfrac{f^{\prime}(x)}{[f(x)]^{n}} d x$ put $f(x)=t$ & proceed.

(2) $\displaystyle \int \dfrac{d x}{a x^{2}+b x+c},$ $\displaystyle \int \dfrac{d x}{\sqrt{a x^{2}+b x+c}},$ $\displaystyle \int \sqrt{a x^{2}+b x+c}\,\,  d x$

Express $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ in the form of perfect square & then apply the standard results.

(3) $\displaystyle \int \dfrac{\mathrm{px}+\mathrm{q}}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx},$ $\displaystyle \int \dfrac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}$

Express $\mathrm{px}+\mathrm{q}$$=\mathrm{A}$ (differential coefficient of quadratic term of denominator ) $+\mathrm{B}$.

(4) $\displaystyle \int e^{x}\left[f(x)+f^{\prime}(x)\right] d x$$=e^{x} \cdot f(x)+c$

(5) $\displaystyle \int\left[f(x)+x f^{\prime}(x)\right] d x$$=x f(x)+c$

(6) $\displaystyle \int \dfrac{d x}{x\left(x^{n}+1\right)} \,\, n \in N$, take $x^{n}$ common & put $1+x^{-n}$$=t$.

(7) $\displaystyle \int \dfrac{d x}{x^{2}\left(x^{n}+1\right)^{(n-1) / n}}\,\, n \in N$, take $x^{n}$ common & put $1+x^{-n}$$=t^{n}$

(8) $\displaystyle \int \dfrac{d x}{x^{n}\left(1+x^{n}\right)^{1 / n}}$, take $x^{n}$ common and put $1+x^{-n}$$=t$.

(9) $\displaystyle \int \dfrac{d x}{a+b \sin ^{2} x} \quad$ OR $\quad \displaystyle \int \dfrac{d x}{a+b \cos ^{2} x}$

OR $\displaystyle \int \dfrac{d x}{a \sin ^{2} x+b \sin x \cos x+c \cos ^{2} x}$

Multiply $\mathrm{N}^{\mathrm{r}}$ & $\mathrm{D}^{\mathrm{r}}$ by $\sec ^{2} \mathrm{x}$ & put $\tan \mathrm{x}$$=\mathrm{t}$

(10) $\displaystyle \int \dfrac{d x}{a+b \sin x}$ OR $\displaystyle \int \dfrac{d x}{a+b \cos x}$ OR $\displaystyle \int \dfrac{d x}{a+b \sin x+c \cos x}$

Convert sines & cosines into their respective tangents of half the angles, put tan $\dfrac{x}{2}$$=t$

(11) $\displaystyle \int \dfrac{a \cdot \cos x+b \cdot \sin x+c}{p \cdot \cos x+q \cdot \sin x+r} \,\, d x$

Express Numerator $\left(\mathrm{N}^{r}\right) \equiv \ell\left(\mathrm{D}^{r}\right)+\mathrm{m} \dfrac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{D}^{r}\right)+\mathrm{n}$ & proceed.

(12) $\displaystyle \int \dfrac{x^{2}+1}{x^{4}+K x^{2}+1} d x$ OR $\displaystyle \int \dfrac{x^{2}-1}{x^{4}+K x^{2}+1} d x$,

where $\mathrm{K}$ is any constant. Divide $N r$ & Dr by $x^{2}$, then put $x-\dfrac{1}{x}$$=t$ OR $x+\dfrac{1}{x}$$=t$ respectively & proceed

(13) $\displaystyle \int \dfrac{d x}{(a x+b) \sqrt{p x+q}}$ & $\displaystyle \int \dfrac{d x}{\left(a x^{2}+b x+c\right) \sqrt{p x+q}} ;$ put $p x+q$$=t^{2}$

(14) $\displaystyle \int \dfrac{d x}{(a x+b) \sqrt{p x^{2}+q x+r}}$, put $a x+b$$=\dfrac{1}{t}$;

$\displaystyle \int \dfrac{d x}{\left(a x^{2}+b x+c\right) \sqrt{p x^{2}+q x+r}}$, put $x$$=\dfrac{1}{t}$

(15) $\displaystyle \int \sqrt{\dfrac{x-\alpha}{\beta-x}}\, d x$ OR $\displaystyle \int \sqrt{(x-\alpha)(\beta-x)} ;$ put $x=\alpha \cos ^{2} \theta+\beta \sin ^{2} \theta$

$\displaystyle \int \sqrt{\dfrac{x-\alpha}{x-\beta}} \, d x$ OR $\displaystyle \int \sqrt{(x-\alpha)(x-\beta)} ;$ put $x=\alpha \sec ^{2} \theta-\beta \tan ^{2} \theta$

$\displaystyle \int \dfrac{\mathrm{dx}}{\sqrt{(x-\alpha)(x-\beta)}} ;$ put $x-\alpha=t^{2}$ or $x-\beta=t^{2}$

(16) To integrate $\displaystyle \int \sin ^{m} x \cos ^{n} x d x$.

(i) If $m$ is odd positive integer put $\cos x$$=t$.

(ii) If $n$ is odd positive integer put $\sin x$$=t$

(iii) If $m+n$ is negative even integer then put $\tan x=t.$

(iv) If $\mathrm{m}$ and $\mathrm{n}$ both even positive integer then use

$\sin ^{2} x$$=\dfrac{1-\cos 2 x}{2}, \cos ^{2} x$$=\dfrac{1+\cos 2 x}{2}$

---

---

(b) Integration by parts:

$\displaystyle \int \mathrm{u} . \mathrm{v}\, \mathrm{dx}$$=\mathrm{u} \displaystyle \int \mathrm{v}\, \mathrm{dx}-\displaystyle \int\left[\dfrac{d \mathrm{u}}{\mathrm{dx} } \cdot \int \mathrm{v}\, \mathrm{dx}\right] \mathrm{dx}$ where $\mathrm{u}$ & $\mathrm{v}$ are differentiable functions.

Note : While using integration by parts, choose u & v such that

(i) $\displaystyle \int \mathrm{v \, dx}$   &   (ii) $\displaystyle \int\left[\dfrac{\mathrm{du}}{\mathrm{dx}} \cdot \displaystyle \int \mathrm{v} \, \mathrm{dx}\right] \mathrm{dx}$ is simple to integrate.

This is generally obtained, by keeping the order of $\mathrm{u}$ & v as per the order of the letters in ILATE, where; I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function.

---

---

(c) Partial fraction :

Rational function is defined as the ratio of two polynomials in the form $\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}$, where $\mathrm{P}(\mathrm{x})$ and $\mathrm{Q}(\mathrm{x})$ are polynomials in $x$ and $Q(x) \neq 0$. If the degree of $P(x)$ is less than the degree of $\mathrm{Q}(\mathrm{x})$, then the rational function is called proper, otherwise, it is called improper. The improper rational function can be reduced to the proper rational functions by long division process. Thus, if $\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}$ is improper, then $\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}$$=\mathrm{T}(\mathrm{x})+\dfrac{\mathrm{P}_{1}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}$, where $\mathrm{T}(\mathrm{x})$ is a polynomial in $\mathrm{x}$ and $\dfrac{\mathrm{P}_{1}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}$ is proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods.

$\begin{array}{|l|l|l|} \hline \text { S. No. } & \begin{array}{l} \text { Form of the } \\ \text { rational function } \end{array} & \begin{array}{l} \text { Form of the } \\ \text { partial fraction } \end{array} \\ \hline \text { 1. } & \frac{\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}}{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})} & \frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{B}}{\mathrm{x}-\mathrm{b}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{c}} \\ \hline \text { 2. } & \frac{\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}}{(\mathrm{x}-\mathrm{a})^{2}(\mathrm{x}-\mathrm{b})} & \frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{B}}{(\mathrm{x}-\mathrm{a})^{2}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{b}} \\ \hline \text { 3. } & \frac{\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}}{(\mathrm{x}-\mathrm{a})\left(\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}\right)} & \frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}} \\ \hline \end{array}$

where $x^2 + bx + c$ cannot be factorised further

Note: 

In competitive exams, partial fraction are generally found by inspection by noting following fact :

$\dfrac{1}{(\mathrm{x}-\alpha)(\mathrm{x}-\beta)}$$=\dfrac{1}{(\alpha-\beta)}\left(\dfrac{1}{\mathrm{x}-\alpha}-\dfrac{1}{\mathrm{x}-\beta}\right)$

It can be applied to the case when $\mathrm{x}^{2}$ or any other function is there in all places of $x$. 

Example : 

(1) $\dfrac{1}{\left(x^{2}+1\right)\left(x^{2}+3\right)}$$=\dfrac{1}{2}\left(\dfrac{1}{t+1}-\dfrac{1}{t+3}\right) \quad$ take $x^{2}=t$ 

(2) $\dfrac{1}{x^{4}\left(x^{2}+1\right)}$$=\dfrac{1}{x^{2}}\left(\dfrac{1}{x^{2}}-\dfrac{1}{x^{2}+1}\right)$$=\dfrac{1}{x^{4}}-\left(\dfrac{1}{x^{2}}-\dfrac{1}{x^{2}+1}\right)$ 

(3) $\dfrac{1}{x^{3}\left(x^{2}+1\right)}$$=\dfrac{1}{x}\left(\dfrac{1}{x^{2}}-\dfrac{1}{x^{2}+1}\right)$$=\dfrac{1}{x^{3}}-\dfrac{1}{x\left(x^{2}+1\right)}$

---

---