x/(x-1)²(x+2) \, dx =

Correct Answer is : −13(1x−1)+29log⁡∣x−1x+2∣+c-1/3 ≤ft( 1/x-1 ) + 2/9 ≤ft| x-1/x+2 | + c−31(x−11)+92logx+2x−1+c

1/2 ≤ft( 1/x-1 ) + 2/9 ≤ft| x-1/x+2 | + c

-1/3 ≤ft( 1/x-1 ) + 2/9 ≤ft| x-1/x+2 | + c

Correct Answer is : −13(1x−1)+29log⁡∣x−1x+2∣+c-1/3 ≤ft( 1/x-1 ) + 2/9 ≤ft| x-1/x+2 | + c−31(x−11)+92logx+2x−1+c

Test Index

AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

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

$\frac{2}{9} \log \left| \frac{x-1}{x+2} \right| + c$
$\frac{2}{9} \log \left| \frac{x+2}{x-1} \right| + c$
$\frac{1}{2} \left( \frac{1}{x-1} \right) + \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| + c$
$-\frac{1}{3} \left( \frac{1}{x-1} \right) + \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| + c$

Solution:

Consider the expression,

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

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

Now by partial fraction,

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

+ \frac{C}{x+2}

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

+ C(x-1)^2

Compare the coefficients,

A = \frac{2}{9}, B = \frac{1}{4}, C = -\frac{2}{9}

So,

\int \frac{x}{x^3-3x+2} \, dx = \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}

= \frac{2}{9} \log|x-1| - \frac{1}{3} \left( \frac{1}{x-1} \right)

- \frac{2}{9} \log|x+2| + c

= -\frac{1}{3} \left( \frac{1}{x-1} \right) + \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| + c

Go to Question:

Prev Question

Next Question