Name: Evaluation of Definite Integrals by Substitution
Uploaded: 2019-07-19
Duration: 11 min 3 s
Description: Did you know that there is a way to evaluate definite integrals by the method of substitution? Watch this exciting video on the concept to learn about it.

Question

Evaluate

\int \frac{1}{x^{3} + 1} d x

Accepted Answer

Let $I = \int \frac{1}{x^{3} + 1} d x$

$\Rightarrow I = \int \frac{1}{(1 + x) (x^{2} - x + 1)} d x$

Consider, $\frac{1}{(1 + x) (x^{2} - x + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} - x + 1}$

$\Rightarrow 1 = A (x^{2} - x + 1) + (B x + C) (x + 1)$

$\Rightarrow 1 = A x^{2} - A x + A + B x^{2} + C x + B x + C$

$\Rightarrow 1 = x^{2} (A + B) + x (- A + B + C) + A + C$

On comparing the coefficients of like powers of $x$ , we get:

$A + B = 0, - A + B + C = 0, A + C = 1$

On solving this, we get:

$A = \frac{1}{3}, B = - \frac{1}{3}, C = \frac{2}{3}$

Thus, $I = \frac{1}{3} \int \frac{1}{x + 1} d x + \int \frac{- \frac{x}{3} + \frac{2}{3}}{x^{2} - x + 1} d x$

$= \frac{1}{3} \log (x + 1) + \frac{1}{3} \int \frac{- x + 2}{x^{2} - x + 1} d x$

$= \frac{1}{3} \log (x + 1) + \frac{1}{6} \int \frac{- 2 x + 4}{x^{2} - x + 1} d x$

$= \frac{1}{3} \log (x + 1) + \frac{1}{6} \int \frac{- 2 x + 1 + 3}{x^{2} - x + 1} d x$

$= \frac{1}{3} \log (x + 1) + \frac{1}{6} [- \int \frac{(2 x - 1)}{x^{2} - x + 1} d x + \int \frac{3}{x^{2} - x + 1} d x]$

Consider, $\int \frac{(2 x - 1)}{x^{2} - x + 1} d x$

Putting $x^{2} - x + 1 = t$

$\Rightarrow (2 x - 1) d x = d t$

Thus, $\int \frac{(2 x - 1)}{x^{2} - x + 1} d x = \int \frac{d t}{t} = \log (t) + c = \log (x^{2} - x + 1) + c$

Now consider, $\int \frac{3}{x^{2} - x + 1} d x$

$= \int \frac{3}{x^{2} - x + \frac{1}{4} - \frac{1}{4} + 1} d x$

$= \int \frac{3}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} d x$

$= 3 \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$

$= 3 \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + k$

$= 2 \sqrt{3} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + k$

Thus, $I = \frac{1}{3} \log (x + 1) - \frac{1}{6} \int \frac{(2 x - 1)}{x^{2} - x + 1} d x + \frac{1}{6} \int \frac{3}{x^{2} - x + 1} d x$

$\Rightarrow I = \frac{1}{3} \log (x + 1) - \frac{1}{6} \log (x^{2} - x + 1) + \frac{1}{6} \cdot 2 \sqrt{3} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) (C = c + k)$

$\Rightarrow I = \frac{1}{3} \log (x + 1) - \frac{1}{6} \log (x^{2} - x + 1) + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C$

Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Manipur Board-2018

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Manipur Board-2018

Questions from Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Manipur Board-2018 with Hints & Solutions

Learn and Prepare for any exam you want !