Question

$\int_{}^{}\frac{x^{3} + 3x + 2}{(x^{2} + 1)^{2}(x + 1)}$dx = $\frac{1}{4}$ log (x² + 1) + $\frac{x}{x^{2} + 1}$ + $\frac{3}{2}$ tan^–1 x – A. Then A is equal to –

• A. $\frac{3}{2}$log |x + 1| + c
• B. $\frac{2}{3}$log |x + 1| + c
• C. $\frac{1}{2}$log |x + 1| + c
• D. $\frac{1}{2}$log |x² + 1| + c

Answer 1

Answer: (C)

$\frac{1}{2}$log |x + 1| + c

Answer 2

Let I = $\int_{}^{}\frac{x^{3} + 3x + 2}{(x^{2} + 1)^{2}(x + 1)}$dx, then

I = dx

= $\int \frac { x } { \left( x ^ { 2 } + 1 \right) ( x + 1 ) }$ dx + 2 $\int_{}^{}\frac{1}{(x^{2} + 1)^{2}}$ dx

= I₁ + 2I₂, where

I₁ = $\int_{}^{}\frac{x}{(x^{2} + 1)(x + 1)}$dx and I₂ = $\int_{}^{}\frac{1}{(x^{2} + 1)^{2}}$dx

Now, I₁ = $\int_{}^{}\frac{x}{(x^{2} + 1)(x + 1)}$dx

= $\int \frac { 1 } { 2 }$ $\left( \frac{x + 1}{x^{2} + 1} - \frac{1}{x + 1} \right)$dx [Using partial fractions]

= $\frac { 1 } { 2 }$ $\int_{}^{}\frac{x}{x^{2} + 1}$dx +$\frac { 1 } { 2 }$ $\int_{}^{}\frac{1}{x^{2} + 1}$dx ––$\frac { 1 } { 2 }$ $\int_{}^{}\frac{1}{x + 1}$dx

= $\frac{1}{4}$log (x² + 1) +$\frac{1}{2}$tan^–1 x – $\frac{1}{2}$log |x + 1| + c₁

and, I₂ = $\int_{}^{}\frac{1}{(x^{2} + 1)^{2}}$dx

= $\frac{1}{2}$ $\frac{x}{(x^{2} + 1)}$+ $\frac{1}{2}$tan^–1 x + c₂

Hence, I = I₁ + 2I₂

= $\frac{1}{4}$log (x² + 1) + $\frac{x}{x^{2} + 1}$ + $\frac{3}{2}$ tan^–1 x – $\frac{1}{2}$log |x + 1| + c.

Hence (3) is the correct answer.