Question

If $\int_{}^{}\frac{2x^{2} + 3}{(x^{2} - 1)(x^{2} + 4)}dx = a\log\left( \frac{x - 1}{x + 1} \right) + b\tan^{- 1}\left( \frac{x}{2} \right) + c$ then values of a and b are

• A. $1, - 1$
• B. $- 1,1$
• C. $1/2, - 1/2$
• D. $1/2,1/2$

Answer 1

Answer: (D)

$1/2,1/2$

Answer 2

Put $x^{2} = y$

$\therefore$ $\frac { 2 x ^ { 2 } + 3 } { \left( x ^ { 2 } - 1 \right) \left( x ^ { 2 } + 4 \right) } = \frac { 2 y + 3 } { ( y - 1 ) ( y + 4 ) }$

$\Rightarrow$ $\frac{2y + 3}{(y - 1)(y + 4)} = \frac{A}{(y - 1)} + \frac{B}{(y + 4)}$

$\therefore$ $2y + 3 = A(y + 4) + B(y - 1)$

Comparing the coefficient of $y$and constant terms

$\Rightarrow$ $A + B = 2$ and $4A - B = 3$

$\therefore A = 1$and $B = 1$

$\therefore$ $I = \int \frac { 1 } { y - 1 } d x + \int \frac { 1 } { y + 4 } d x$

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

$\Rightarrow$ $I = \frac { 1 } { 2 } \log \left| \frac { x - 1 } { x + 1 } \right| + \frac { 1 } { 2 } \tan ^ { - 1 } \frac { x } { 2 } + c$

$\therefore$ $a = \frac { 1 } { 2 }$ and $b = \frac { 1 } { 2 }$