# Integrals of Some Particular Function

We have already discussed about the Integration of functions, Methods of Integration, Integration of Trigonometric functions, Integration of Inverse trigonometric function etc.

Here we will discuss integration of some of the important function and see their use in many other standard integrals:

 S.No Integral function Integral value 1 $\int \frac{dx}{x^{2}- a^{2}}$ $\frac{1}{2a} \log \left | \frac{x-a}{x+a} \right | + C$ 2 $\int \frac{dx}{a^{2}- x^{2}}$ $\frac{1}{2a} \log \left | \frac{a+x}{a-x} \right | + C$ 3 $\int \frac{dx}{x^{2} + a^{2}}$ $\frac{1}{a}\tan^{-1}\left ( \frac{x}{a} \right ) + C$ 4 $\int \frac{dx}{\sqrt{x^{2} – a^{2}}}$ $\log \left | x + \sqrt{x^{2}- a^{2}} \right | + C$ 5 $\int \frac{dx}{\sqrt{a^{2} – x^{2}}}$ $\sin^{-1}\left ( \frac{x}{a} \right ) + C$ 6 $\int \frac{dx}{\sqrt{x^{2} + a^{2}}}$ $\log \left | x + \sqrt{x^{2} + a^{2}} \right | + C$

Let us now prove the integration of particular functions:

(1) $\large \mathbf{\int \frac{dx}{x^{2} – a^{2}} = \frac{1}{2a} \log \left | \frac{x-a}{x+a} \right | + C }$

The integral function can be splitted into the sums of partial fraction, i.e.

$\int \frac{dx}{x^{2}- a^{2}} = \int \frac{dx}{(x-a)(x+a)}= \int \frac{A}{(x-a)}.dx + \int \frac{B}{(x+a)}.dx$……………(i)

Solving for values of A and B, we have,

$1= A(x+a) + B(x-a)$,

Putting x = a and then -a, we get the values of A and B to be $\frac{1}{2a}$ and $-\frac{1}{2a}$ respectively.

Substituting these values in (i), we have

$\int \frac{dx}{x^{2}- a^{2}} = \int \frac{dx}{2a(x-a)} + \int \frac{-dx}{2a(x+a)}$

$= \frac{1}{2a} \left [ \int \frac{dx}{(x-a)} – \int \frac{dx}{(x+a)} \right ]$

$= \frac{1}{2a} \left [ \log \left | x – a \right | – \log \left | x + a \right | \right ] + C$

$= \frac{1}{2a} \log \left | \frac{x – a}{x+a} \right | + C$

(2) $\large \mathbf{\int \frac{dx}{a^{2}- x^{2}} = \frac{1}{2a} \log \left | \frac{a+x}{a-x} \right | + C }$

Breaking function into the sums of partial fraction, we have

$\int \frac{dx}{a^{2}- x^{2}} = \int \frac{A}{a-x}.dx + \int \frac{B}{a+x}.dx$

Solving for values of A and B, we have

$\int \frac{dx}{a^{2}- x^{2}} = \int \frac{dx}{2a(a-x)} + \int \frac{dx}{2a(a+x)}$

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

$=\frac{1}{2a} \left [ -\log \left | a-x \right | + \log \left | a+x \right | \right ] + C$

$=\frac{1}{2a} \log \left | \frac{a+x}{a-x} \right | + C$

(3) $\large \mathbf{\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a}\tan^{-1}\left ( \frac{x}{a} \right ) + C}$

Substituting $x = a\tan \theta$……….(i)

$dx = a\sec^{2} \theta .d\theta$

$\int \frac{dx}{x^{2} + a^{2}} = \int \frac{a \sec^{2}\theta .d\theta }{a^{2}\tan^{2}\theta + a^{2}}$

$= \frac{1}{a}\int \frac{\sec^{2}\theta .d\theta }{\ \sec^{2}\theta } = \frac{1}{a}\int d\theta$

$= \frac{1}{a} \theta + C$ ………………(ii)

From (i), we know $\theta = \tan^{-1}\frac{x}{a}$,

therefore $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a}\tan^{-1} \frac{x}{a} + C$

(4) $\large \mathbf{\int \frac{dx}{\sqrt{x^{2} – a^{2}}} = \log \left | x + \sqrt{x^{2}- a^{2}} \right | + C}$

Substituting $x = a\sec \theta$ ……….(i)

$dx = a\sec \theta \tan \theta .d\theta$

$\int \frac{dx}{\sqrt{x^{2} – a^{2}}} = \int \frac{a\sec \theta \tan \theta .d\theta}{\sqrt{a^{2}\sec^{2}\theta – a^{2}}}$

$= \int \frac{a\sec \theta \tan \theta .d\theta}{a \sqrt{\tan^{2}\theta }}$

$= \int \sec \theta . d\theta$

$= \log \left | \sec \theta + \tan \theta \right | + C_{1}$……….(ii)

from (i) we know $\sec \theta = \frac{x}{a}$ and $\tan \theta = \sqrt{\sec^{2} \theta – 1 } = \sqrt{ \frac{x^{2}}{a^{2}} – 1}$

Substituting these values in equation (ii), we have

$\int \frac{dx}{\sqrt{x^{2} – a^{2}}} = \log \left | \frac{x}{a} + \sqrt{ \frac{x^{2}}{a^{2}}-1}\right | + C_{1}$

$= \log \left | \frac{x + \sqrt{x^{2}- a^{2}}}{a} \right | + C_{1}$

$= \log \left | x + \sqrt{x^{2}- a^{2}} \right | – \log \left | a \right |+ C_{1}$

$= \log \left | x + \sqrt{x^{2}- a^{2}} \right | + C$,(where $C = C_{1} + \log \left | a \right |$ )

(5) $\large \mathbf{\int \frac{dx}{\sqrt{a^{2} – x^{2}}} = \sin^{-1}\left ( \frac{x}{a} \right ) + C}$

Putting $x = a \sin \theta$

$dx = a \cos \theta .d \theta$

$\int \frac{dx}{\sqrt{a^{2} – x^{2}}} = \int \frac{a \cos \theta. d\theta}{\sqrt{a^{2}- a^{2}\sin^{2}\theta}}$

$= \int \frac{a \cos \theta. d\theta}{a \sqrt{1- \sin^{2}\theta}}$

$= \int \frac{a \cos \theta. d\theta}{a \cos\theta} = \int d\theta$

$= \theta + C$

$= \sin^{-1}\frac{x}{a} + C$

(6) $\large \mathbf{\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \log \left | x + \sqrt{x^{2} + a^{2}} \right | + C}$

Putting $x = a \tan \theta$…………….(i)

$dx = a \sec^{2} \theta d\theta$

$\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \int \frac{a \sec^{2} \theta d\theta}{\sqrt{a^{2}\tan^{2}\theta + a^{2}}}$

$= \int \frac{a \sec^{2} \theta d\theta}{a\sqrt{\tan^{2}\theta + 1}} = \int \frac{a \sec^{2} \theta d\theta}{a\sqrt{\sec^{2}\theta}}$

$= \int \sec \theta.d\theta$

$= \log \left | \sec \theta + \tan \theta \right | + c$……………….(ii)

From (i), we have $\tan \theta = \frac{x}{a}$ and $\sec \theta = \sqrt{\tan^{2}\theta + 1} = \sqrt{\frac{x^{2}}{a^{2}}+1}$

Putting these value in (ii), we get

$\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \log \left | \frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}}+1} \right | + c$

$= \log \left | \frac{x+ \sqrt{x^{2}+1}}{a} \right | + c$

$= \log \left | x+ \sqrt{x^{2}+1} \right | – \log a + c$

$= \log \left | x+ \sqrt{x^{2}+1} \right | + C$  (where $C = c – \log a$)

These standard formulae can be used to obtain new formulae and can be applied directly to evaluate other integrals.

(7) $\large \mathbf{\int \frac{dx}{ax^{2} + bx + c}}$

the denominator can be written as $ax^{2} + bx + c = a\left [ x^{2} + \frac{b}{a}x + \frac{c}{a} \right ] = a\left [ \left ( x + \frac{b}{2a} \right )^{2} + \left ( \frac{c}{a} – \frac{b^{2}}{4a^{2}} \right )\right ]$

Substituting $x + \frac{b}{2a} = t$, so $dx = dt$

Also $\frac{c}{a} – \frac{b^{2}}{4a^{2}} = \pm k^{2}$

Hence the integral becomes,

$\int \frac{dx}{ax^{2} + bx + c} = \frac{1}{a}\int \frac{dt}{t^{2}\pm k^{2}}$

(8) $\large \mathbf{{\int \frac{dx}{\sqrt{ax^{2} + bx + c}}}}$

Similarly like equation (7), we can obtain the standard integral as,

$\int \frac{dx}{\sqrt{ax^{2} + bx + c}}= \frac{1}{a}\int \frac{dt}{\sqrt{t^{2}\pm k^{2}}}$

(9) $\large \mathbf{\int \frac{px+q}{ax^{2} + bx + c}.dx}$,

where p,q,a,b,c are constants

$px + q = A \frac{\mathrm{d} }{\mathrm{d} x}(ax^{2}+bx+c) = A (2ax+ b)+B$

we equate the coefficient of x of both the sides to determine the value of A and B, and hence the integral is reduced to one of the known forms.

(10) $\large \mathbf{\int \frac{px+q}{\sqrt{ax^{2} + bx + c}}.dx}$

Similarly like equation (9), we can obtain the standard form.

 Example: Find the Integral of the function $\large \int \frac{dx}{\sqrt{7x^{2}-2x}}$. Solution:The given function can be converted into the standard form $\large \int \frac{dx}{\sqrt{7x^{2}-2x}} = \int \frac{dx}{\sqrt{7}.\sqrt{x^{2}-\frac{2}{7}x}}$ $\large = \frac{1}{\sqrt{7}}\int \frac{dx}{\sqrt{\left ( x- \frac{1}{7} \right )^{2}-\left ( \frac{1}{7} \right )^{2}}}$(completing the squares) Substituting $\large x – \frac{1}{7} = t$, so dx = dt therefore $\large \int \frac{dx}{\sqrt{7x^{2}-2x}} = \frac{1}{\sqrt{7}} \int \frac{dt}{\sqrt{t^{2}- \left ( \frac{1}{7} \right )^{2}}}$ $\large = \frac{1}{\sqrt{7}} \log \left | t + \sqrt{t^{2}- \left ( \frac{1}{7} \right )^{2}} \right | + C$ $\large = \frac{1}{\sqrt{7}} \log \left | x- \frac{1}{7} + \sqrt{\left ( x- \frac{1}{7} \right )^{2}- \left ( \frac{1}{7} \right )^{2}} \right | + C$ $\large = \frac{1}{\sqrt{7}} \log \left | x- \frac{1}{7} + \sqrt{x^{2}-\frac{2}{7}x} \right | + C$<

#### Practise This Question

Which of the following represents a line graph for the given histogram?