Algebraic Integrals

Finding the solution of algebraic integrals is a lengthy and complex process. Hence, a standard method is to be followed to attain the solution in an easy way. In this section, we will discuss different forms of algebraic integrals and tricks given will help aspirants to solve all the problems at their own pace.

The types of integrals are as follows:

  • The integral can be called a definite integral if the function f(x) is between the limits ‘a’ and ‘b’, where d/dx (F(x)) = f(x).
It is represented as follows:

abf(x)\int_{a}^{b}f(x) dx = F(b) – F(a)
  • Indefinite integrals refer to the evaluation of the indefinite area. The diagram below explains the difference between definite and indefinite integral.

Difference between definite and indefinite integrals

Algebraic Integrals Forms

The following forms of integrals can be categorised into algebraic integrals.

1] 1ax2+bx+cdx\int \frac{1}{ax^2+bx+c}dx or reducible to 1ax2+bx+cdx\int \frac{1}{ax^2+bx+c}dx

Tricks to Solve: In these integrals try expressing the term ax2 + bx + c as sum or difference of the two squares. The following steps can be followed:

  • By dividing or multiplying throughout, the coefficient of x2 can be made unity.
  • To express it in the form [x+b2a]2[x+\frac{b}{2a}]^2 + 4acb24a2\frac{4ac-b^2}{4a^2}, add and subtract the square of half of the coefficient of x.
  • This reduces to into one of the forms and can be easily integrated. 1x2a2 or 1x2+a2 or 1a2x2\frac{1}{x^2-a^2}\ or\ \frac{1}{x^2+a^2}\ or\ \frac{1}{a^2-x^2}

2] 1ax2+bx+cdx\int \frac{1}{\sqrt{ax^2+bx+c}}dx or reducible to 1ax2+bx+cdx\int \frac{1}{\sqrt{ax^2+bx+c}}dx

Tricks to Solve: For integrating these functions the following steps can be followed.

  • Make the coefficient of x2 as unity.
  • To express the term within the square root in the form [x+b2a]2[x+\frac{b}{2a}]^2 + 4acb24a2\frac{4ac-b^2}{4a^2}, Add and subtract the square of half of the coefficient of x.
  • This reduces to into one of the forms and can be easily integrated. 1x2a2 or 1x2+a2 or 1a2x2\frac{1}{x^2-a^2}\ or\ \frac{1}{x^2+a^2}\ or\ \frac{1}{a^2-x^2}

3] px+qax2+bx+c\int \frac{px+q}{ax^2+bx+c}

Tricks to Solve:

  • In such cases, the numerator can be expressed as the sum of constant times differentiation of denominator and a constant that is, px + q is written as A (2ax + b) + C, where 2ax + b= d/dx (ax2+bx+c). A and C are arbitrary constants.
  • By equating on the coefficients of like powers on both sides, the values of the constants can be easily obtained.
  • Replace the numerator (px + q) by A (2ax + b) + C in the given integral which will give, px+qax2+bx+c=A2ax+bax2+bx+cdx+C1ax2+bx+cdx\int\frac{px+q}{ax^2+bx+c}=A\int\frac{2ax+b}{ax^2+bx+c}dx+C\int\frac{1}{ax^2+bx+c}dx
  • The right hand side value thus obtained in the previous step can be integrated and then constant values can be substituted.

4] px+qax2+bx+c\int\frac{px+q}{\sqrt{ax^2+bx+c}}

Tricks to Solve:

In this case, we proceed as follows:

  • The numerator is expressed as the sum of constant times differentiation of denominator and a constant, that is, px + q is written as A (2ax + b) + C, where 2ax + b=d/dx (ax2+bx+c). A and C are arbitrary constants.
  • The values of constants are computed by equating the coefficients of like powers on both sides.
  • Replace the numerator (px + q) by A (2ax + b) + C in the given integral which will give, px+qax2+bx+cdx=A2ax+bax2+bx+cdx+C1ax2+bx+cdx\int\frac{px+q}{\sqrt{ax^2+bx+c}}dx=A\int\frac{2ax+b}{\sqrt{ax^2+bx+c}}dx+C\int \frac{1}{\sqrt{ax^2+bx+c}}dx
  • The right hand side value thus obtained in the previous step can be integrated and then constant values can be substituted.

5] ax2+bx+cdx\int \sqrt{ax^2+bx+c}dx

Tricks to Solve:

In such type of integrals, the following steps are followed:

  • Make the coefficient of x2 as unity.
  • To obtain the form [x+b2a]2[x+\frac{b}{2a}]^2 + 4acb24a2\frac{4ac-b^2}{4a^2}, Add and subtract the square of half of the coefficient of x.
  • Henceforth the integral reduces to, \int \sqrt{a^2+x^2}dx,\ or \  \int \sqrt{a^2-x^2}dx\ or \ \int \sqrt{x^2-a^2}dx
  • These integrals can be computed easily by using direct formulae.

Also read

Definite and indefinite integration

Integration of functions

Applications of integration

Algebraic Integrals Examples

Below are some solved examples on algebraic integrals for IIT JEE aspirants.

Example 1: x+12x2+x3dx\int \frac{x+1}{\sqrt{2x^2+x-3}}dx

Solution:

By completing the square,

Algebraic Integrals Example

Algebraic Integrals Example for IIT JEE

Example 2: 1x2+6x+5dx\int \frac{1}{x^2+6x+5}dx

Solution:

Algebraic Expression Problem

By completing the square,

Example on Algebraic Integrals

Example 3: 16x25x+1dx\int \frac{1}{\sqrt{6x^2-5x+1}}dx

Solution:

Example on Algebraic Integrals for IIT JEE

By completing the square,

Solved Problem on Algebraic Integrals

Solved Problems on Algebraic Integrals

Example 4: I=010sinxdx I=\int _0^{10}\left|sinx\right|\:dx\\

Solution: 

=0πsin(x)dx+π2πsin(x)dx+2π3πsin(x)dx+3π10sin(x)dx0πsin(x)dx=[cos(x)]0π=2π2πsin(x)dx=π2πsin(x)dx=[cos(x)]π2π=(2)2π3πsin(x)dx=[cos(x)]2π3π=23π10sin(x)dx=[cos(x)]3π10=(cos(10)1)=2+2+2+cos(10)+1=cos(10)+7=\int _0^{\pi }\sin \left(x\right)dx+\int _{\pi }^{2\pi }-\sin \left(x\right)dx+\int _{2\pi }^{3\pi }\sin \left(x\right)dx+\int _{3\pi }^{10}-\sin \left(x\right)dx\\\int _0^{\pi }\sin \left(x\right)dx=\left[-\cos \left(x\right)\right]^{\pi }_0=2\\ \int _{\pi }^{2\pi }-\sin \left(x\right)dx=-\int _{\pi }^{2\pi }\sin \left(x\right)dx=-\left[-\cos \left(x\right)\right]^{2\pi }_{\pi }=-\left(-2\right)\\\int _{2\pi }^{3\pi }\sin \left(x\right)dx=\left[-\cos \left(x\right)\right]^{3\pi }_{2\pi }=2\\\int _{3\pi }^{10}-\sin \left(x\right)dx=-\left[-\cos \left(x\right)\right]^{10}_{3\pi }=-\left(-\cos \left(10\right)-1\right)\\=2+2+2+\cos \left(10\right)+1\\=\cos \left(10\right)+7

Example 5: If In=0π/4tannxdxI_{n}=\int_{0}^{\pi/4}tan^{n}xdx, then limnn[In+In+2]=\lim_{n\rightarrow \infty}n[I_{n}+I_{n+2}]=

Solution:

In+In+2=0π/4tannx[1+tan2xdx]0π/4tannxsec2xdx=tann+1xn+1 from 0 to π4=10n+1=1n+1In+In+2=1n+1limxn[In+In+2]limxn1n+1=nn+1=nn[1+1n]=1I_{n}+I_{n+2}=\int_{0}^{\pi/4}tan^{n}x[1+tan^{2}xdx]\\ \int_{0}^{\pi/4}tan^{n}xsec^{2}xdx=\frac{tan^{n+1}x}{n+1} \text \ from \ 0 \ to \ \frac{\pi}{4}\\ =\frac{1-0}{n+1}=\frac{1}{n+1}\\ I_{n}+I_{n+2}=\frac{1}{n+1}\\\Rightarrow \lim_{x\rightarrow \infty }n[I_{n}+I_{n+2}]\\ \Rightarrow \lim_{x\rightarrow \infty }n*\frac{1}{n+1}=\frac{n}{n+1}=\frac{n}{n*[1+\frac{1}{n}]}=1

Example 6: 011+7x3dx\int _0^1\sqrt[3]{1+7x}dx

Solution:

Applyusubstitution:u=1+7x=18u37du=1718u3du=1718u13du=17[u13+113+1]18=1[u13+113+1]187u13+113+1=3u434=[u4334]187=17[u4334]18=17454=4528\mathrm{Apply\:u-substitution:}\:u=1+7x\\=\int _1^8\frac{\sqrt[3]{u}}{7}du\\=\frac{1}{7}\cdot \int _1^8\sqrt[3]{u}du\\=\frac{1}{7}\cdot \int _1^8u^{\frac{1}{3}}du\\=\frac{1}{7}\left[\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]^8_1\\=\frac{1\cdot \:\left[\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]^8_1}{7}\\\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}=\frac{3u^{\frac{4}{3}}}{4}\\=\frac{\left[\frac{u^{\frac{4}{3}}\cdot \:3}{4}\right]^8_1}{7}\\=\frac{1}{7}\left[\frac{u^{\frac{4}{3}}\cdot \:3}{4}\right]^8_1\\=\frac{1}{7}\cdot \frac{45}{4}\\=\frac{45}{28}

Example 7: exex+exdx\int \frac{e^x}{e^x+e^{-x}}dx

Solution: 

exex+exdxApplyusubstitution:u=ex=uu2+1duApplyusubstitution:v=u2+1=12vdv12lnv=12ln(ex)2+1=12lne2x+1=12lne2x+1+C\int \frac{e^x}{e^x+e^{-x}}dx\\ \mathrm{Apply\:u-substitution:}\:u=e^x\\ =\int \frac{u}{u^2+1}du\\ \mathrm{Apply\:u-substitution:}\:v=u^2+1\\=\int \frac{1}{2v}dv\\\frac{1}{2}\ln \left|v\right|\\=\frac{1}{2}\ln \left|\left(e^x\right)^2+1\right|\\=\frac{1}{2}\ln \left|e^{2x}+1\right|\\=\frac{1}{2}\ln \left|e^{2x}+1\right|+C