Integral of Particular Functions (With Proof and Example) | Class 12 Math Notes Study Material Download Free PDF

Integral is the method to sum the functions on a large scale. Hence, here we have discussed integrals of some particular functions which are usually used for calculations. These integrals have huge applications in real life, such as finding the area between curves, volume, the average value of the function, kinetic energy, center of mass, work-done, etc.

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

In this article, you will learn the integration of some of the important functions and see their use in many other standard integrals.

Table of Contents

• Integrals of Some Particular Functions
• Proofs of Integrals Functions
  • Integral of function 1
  • Integral of function 2
  • Integral of function 3
  • Integral of function 4
  • Integral of function 5
  • Integral of function 6
• Integral of Some More Functions
  • Integral of function 7
  • Integral of function 8
• Example Based on Integral of Function
  • Integrals of Some Functions Questions

Integrals of Some Particular Functions

• ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
• ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
• ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
• ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
• ∫1/√(a2 – x2) dx = sin-1(x/a) + C
• ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C

These are tabulated below along with the meaning of each part.

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

Also, see:

Proofs of Integrals Functions

Let us now prove the integration of particular functions.

Integral of function 1

(1)

$\begin{array}{l}\int \frac{\mathbf{dx}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{–}{\mathbf{a}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{a}}\log \left|\frac{\mathbf{x}\mathbf{-}\mathbf{a}}{\mathbf{x}\mathbf{+}\mathbf{a}}\right|\mathbf{+}\mathbf{C}\end{array}$

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

$\begin{array}{l}\int \frac{dx}{x^2-a^2}=\int \frac{dx}{(x-a)(x+a)}=\int \frac{A}{(x-a)}.dx+\int \frac{B}{(x+a)}.dx\end{array}$

……………(i)

Solving for values of A and B, we have,

$\begin{array}{l}1=A(x+a)+B(x-a)\end{array}$

,

Putting x = a and then -a, we get the values of A and B to be

$\begin{array}{l}\frac{1}{2a}\end{array}$

and

$\begin{array}{l}-\frac{1}{2a}\end{array}$

respectively.

Substituting these values in (i), we have

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

$\begin{array}{l}=\frac{1}{2a}[\int \frac{dx}{(x-a)}–\int \frac{dx}{(x+a)}]\end{array}$

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

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

Integral of function 2

(2)

$\begin{array}{l}\int \frac{\mathbf{dx}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{a}}\log \left|\frac{\mathbf{a}\mathbf{+}\mathbf{x}}{\mathbf{a}\mathbf{-}\mathbf{x}}\right|\mathbf{+}\mathbf{C}\end{array}$

Breaking function into the sums of partial fraction, we have

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

Solving for values of A and B, we have

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

$\begin{array}{l}=\frac{1}{2a}[\int \frac{dx}{(a-x)}+\int \frac{dx}{(a+x)}]\end{array}$

$\begin{array}{l}=\frac{1}{2a}[-\log |a-x|+\log |a+x|]+C\end{array}$

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

Integral of function 3

(3)

$\begin{array}{l}\int \frac{\mathbf{dx}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{a}}{\tan }^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{x}}{\mathbf{a}}\right)\mathbf{+}\mathbf{C}\end{array}$

Substituting

$\begin{array}{l}x=a\tan \theta \end{array}$

……….(i)

$\begin{array}{l}dx=a{\sec }^2\theta .d\theta \end{array}$

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

$\begin{array}{l}=\frac{1}{a}\int \frac{{\sec }^2\theta .d\theta }{{\sec }^2\theta }=\frac{1}{a}\int d\theta \end{array}$

$\begin{array}{l}=\frac{1}{a}\theta +C\end{array}$

………………(ii)

From (i), we know

$\begin{array}{l}\theta ={\tan }^{-1}\frac{x}{a}\end{array}$

,

therefore

$\begin{array}{l}\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\end{array}$

Integral of function 4

(4)

$\begin{array}{l}\int \frac{\mathbf{dx}}{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{–}{\mathbf{a}}^{\mathbf{2}}}}\mathbf{=}\log |\mathbf{x}\mathbf{+}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}}|\mathbf{+}\mathbf{C}\end{array}$

Substituting

$\begin{array}{l}x=a\sec \theta \end{array}$

 ……….(i)

$\begin{array}{l}dx=a\sec \theta \tan \theta .d\theta \end{array}$

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

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

$\begin{array}{l}=\int \sec \theta .d\theta \end{array}$

$\begin{array}{l}=\log |\sec \theta +\tan \theta |+C_1\end{array}$

……….(ii)

from (i) we know

$\begin{array}{l}\sec \theta =\frac{x}{a}\end{array}$

and

$\begin{array}{l}\tan \theta =\sqrt{{\sec }^2\theta –1}=\sqrt{\frac{x^2}{a^2}–1}\end{array}$

Substituting these values in equation (ii), we have

$\begin{array}{l}\int \frac{dx}{\sqrt{x^2–a^2}}=\log |\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}|+C_1\end{array}$

$\begin{array}{l}=\log \left|\frac{x+\sqrt{x^2-a^2}}{a}\right|+C_1\end{array}$

$\begin{array}{l}=\log |x+\sqrt{x^2-a^2}|–\log \left|a\right|+C_1\end{array}$

$\begin{array}{l}=\log |x+\sqrt{x^2-a^2}|+C\end{array}$

,(where

$\begin{array}{l}C=C_1+\log \left|a\right|\end{array}$

)

Integral of function 5

(5)

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

Putting

$\begin{array}{l}x=a\sin \theta \end{array}$

$\begin{array}{l}dx=a\cos \theta .d\theta \end{array}$

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

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

$\begin{array}{l}=\int \frac{a\cos \theta .d\theta }{a\cos \theta }=\int d\theta \end{array}$

$\begin{array}{l}=\theta +C\end{array}$

$\begin{array}{l}={\sin }^{-1}\frac{x}{a}+C\end{array}$

Integral of function 6

(6)

$\begin{array}{l}\int \frac{\mathbf{dx}}{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}}}\mathbf{=}\log |\mathbf{x}\mathbf{+}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}}|\mathbf{+}\mathbf{C}\end{array}$

Putting

$\begin{array}{l}x=a\tan \theta \end{array}$

…………….(i)

$\begin{array}{l}dx=a{\sec }^2\theta d\theta \end{array}$

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

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

$\begin{array}{l}=\int \sec \theta .d\theta \end{array}$

$\begin{array}{l}=\log |\sec \theta +\tan \theta |+c\end{array}$

……………….(ii)

From (i), we have

$\begin{array}{l}\tan \theta =\frac{x}{a}\end{array}$

and

$\begin{array}{l}\sec \theta =\sqrt{{\tan }^2\theta +1}=\sqrt{\frac{x^2}{a^2}+1}\end{array}$

Putting these value in (ii), we get

$\begin{array}{l}\int \frac{dx}{\sqrt{x^2+a^2}}=\log |\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}|+c\end{array}$

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

$\begin{array}{l}=\log |x+\sqrt{x^2+1}|–\log a+c\end{array}$

$\begin{array}{l}=\log |x+\sqrt{x^2+1}|+C\end{array}$

 (where

$\begin{array}{l}C=c–\log a\end{array}$

)

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

Integral of Some More Functions

We have already discussed the integrals of the first six functions with proofs. Let us see integral of some more functions.

Integral of function 7

(7)

$\begin{array}{l}\int \frac{\mathbf{dx}}{{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}}\end{array}$

the denominator can be written as

$\begin{array}{l}a{x}^2+bx+c=a[x^2+\frac{b}{a}x+\frac{c}{a}]=a[{(x+\frac{b}{2a})}^2+\left(\frac{c}{a}–\frac{b^2}{4a^2}\right)]\end{array}$

Substituting

$\begin{array}{l}x+\frac{b}{2a}=t\end{array}$

, so

$\begin{array}{l}dx=dt\end{array}$

Also

$\begin{array}{l}\frac{c}{a}–\frac{b^2}{4a^2}=\pm k^2\end{array}$

Hence the integral becomes,

$\begin{array}{l}\int \frac{dx}{a{x}^2+bx+c}=\frac{1}{a}\int \frac{dt}{t^2\pm k^2}\end{array}$

Integral of function 8

(8)

$\begin{array}{l}\int \frac{\mathbf{px}\mathbf{+}\mathbf{q}}{{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}}\mathbf{.}\mathbf{dx}\end{array}$

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

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

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.

Example Based on Integral of Function

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

Go through the another example given below:

Example 2: Find the integral of given function ∫x-3(x + 1)dx Solution: Consider the function,  f(x) = x-3(x + 1)  = (x + 1)/x3 = (x/x3) + (1/x3) = (1/x2) + (1/x3) Now, ∫x-3(x + 1)dx = ∫ (1/x2) + (1/x3) dx = (-1/x) + (-1/2x2) + C = -(1/x) – (1/2x2) + C Therefore, ∫x-3(x + 1)dx = -(1/x) – (1/2x2) + C

Integrals of Some Functions Questions

1. Find the integral of the given function: ∫dx/(3x2 + 13x – 10)
2. Find the integral of the given function: ∫dx/√(2x – x2)
3. Evaluate: √(x + 3)/√(5 – 4x + x2) dx