{x d x}{(x-1)²(x+2)}=

Correct Answer is : 29log⁡(x−1)−13×1x−1−29log⁡(x+2)+c2/9 (x-1)-1/3 × 1/x-1-2/9 (x+2)+c92log(x−1)−31×x−11−92log(x+2)+c, where c is a constant of integration

2/9 (x-1)+1/3 × 1/x-1+2/9 (x+2)+c, where c is a constant of integration

2/9 (x-1)-1/3 × 1/(x-1)+2/9 (x+2)+c, where c is a constant of integration

2/9 (x-1)+1/3 × 1/x-1-2/9 (x+2)+c, where c is a constant of integration

2/9 (x-1)-1/3 × 1/x-1-2/9 (x+2)+c, where c is a constant of integration

{x d x}{(x-1)²(x+2)}=

Correct Answer is : 29log⁡(x−1)−13×1x−1−29log⁡(x+2)+c2/9 (x-1)-1/3 × 1/x-1-2/9 (x+2)+c92log(x−1)−31×x−11−92log(x+2)+c, where c is a constant of integration

Test Index

MHTCET 11 May 2024 Shift 1 Paper

Question : 101 of 150

Marks: +1, -0

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

$\frac{2}{9} \log (x-1)+\frac{1}{3} \times \frac{1}{x-1}+\frac{2}{9} \log (x+2)+c$ , where c is a constant of integration
$\frac{2}{9} \log (x-1)-\frac{1}{3} \times \frac{1}{(x-1)}+\frac{2}{9} \log (x+2)+c$ , where c is a constant of integration
$\frac{2}{9} \log (x-1)+\frac{1}{3} \times \frac{1}{x-1}-\frac{2}{9} \log (x+2)+c$ , where c is a constant of integration
$\frac{2}{9} \log (x-1)-\frac{1}{3} \times \frac{1}{x-1}-\frac{2}{9} \log (x+2)+c$ , where c is a constant of integration

Validate

Solution:

Concept:

Use partial fractions to split the integrand, then integrate each simple term separately.

Explanation:

Let

I = \int \frac{x}{(x-1)^2(x+2)}\,dx

Write the integrand as:

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

Multiplying by

(x-1)^2(x+2)

gives:

x = A(x-1)(x+2) + B(x+2) + C(x-1)^2

Put

x = 1

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

Put

x = -2

-2 = 9C \Rightarrow C = -\frac{2}{9}

Equate coefficients of

x^2

A + C = 0 \Rightarrow A = \frac{2}{9}

Substitute

A

B

C

and integrate:

\frac{2}{9}\int \frac{dx}{x-1} + \frac{1}{3}\int \frac{dx}{(x-1)^2} - \frac{2}{9}\int \frac{dx}{x+2}

This gives:

\frac{2}{9}\ln|x-1| - \frac{1}{3}\cdot \frac{1}{x-1} - \frac{2}{9}\ln|x+2| + c

Answer:

Option D:

\frac{2}{9} \log (x-1)-\frac{1}{3} \times \frac{1}{x-1}-\frac{2}{9} \log (x+2)+c

, where

c

is a constant of integration.

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