Name: Integral of Type ((px+q)/(ax^2+bx+c)) dx
Uploaded: 2019-07-19
Duration: 13 min 31 s
Description: In the previous videos, we saw the method of partial fraction functions to integrate. This question can be solved using the same concept. Let us watch the vi...

Question

Evaluate

\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x

Accepted Answer

Given, $\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$

Since the denominator of the integrand contains three non-repeated linear factors. So, we can write the integrand into the sum of three partial fractions. Thus, we have

$\frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} = \frac{A}{(x - 1)} + \frac{B}{x + 2} + \frac{C}{x - 3} . . . . (i)$ , where $A, B,$ and $C$ are real numbers.

$\Rightarrow \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} = \frac{A (x + 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x + 2)}{(x - 1) (x + 2) (x - 3)}$

$\Rightarrow 2 x - 1 = A (x + 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x + 2) . . . . (i i)$

Substituting $x = 3$ in equation $(i i)$ , we have

$C (2) (5) = 2 (3) - 1 = 5$

$\Rightarrow C = \frac{1}{2}$

Now, Substituting $x = - 2$ in equation $(i i)$ , we have

$2 (- 2) - 1 = B (- 3) (- 5)$

$\Rightarrow - 4 - 1 = 15 B$

$\Rightarrow B = - \frac{1}{3}$

Now, Substituting $x = 1$ in equation $(i i)$ , we have

$\Rightarrow 2 (1) - 1 = A (1 + 2) (1 - 3)$

$\Rightarrow 1 = - 6 A$

$\Rightarrow A = - \frac{1}{6}$

Thus, from equation $(i)$ , we have

$\frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} = - \frac{1}{6 (x - 1)} - \frac{1}{3 (x + 2)} + \frac{1}{2 (x - 3)}$

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

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

$= - \frac{1}{6} \log |x - 1| - \frac{1}{3} \log |x + 2| + \frac{1}{2} \log |x - 3| + C$ , where $C$ is an integral constant.

Hence, $\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x = = - \frac{1}{6} \log |x - 1| - \frac{1}{3} \log |x + 2| + \frac{1}{2} \log |x - 3| + C$ , where $C$ is an integral constant.

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

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

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

Learn and Prepare for any exam you want !