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Question: \[\int_{}^{}{\frac{x^{4}}{(x - 1)(x^{2} + 1)}dx =}\]...

∫x4(x−1)(x2+1)dx=\int_{}^{}{\frac{x^{4}}{(x - 1)(x^{2} + 1)}dx =}

A

x(x+2)2+log⁡(x−1)2−log⁡(x2+1)4−tan⁡−1x2+c\frac{x(x + 2)}{2} + \frac{\log(x - 1)}{2} - \frac{\log(x^{2} + 1)}{4} - \frac{\tan^{- 1}x}{2} + c

B

x(x+2)2+log⁡(x−1)2+log⁡(x2+1)4−tan⁡−1x2+c\frac{x(x + 2)}{2} + \frac{\log(x - 1)}{2} + \frac{\log(x^{2} + 1)}{4} - \frac{\tan^{- 1}x}{2} + c

C

x(x+2)2+log⁡(x−1)2+log⁡(x2+1)4+tan⁡−1x2+c\frac{x(x + 2)}{2} + \frac{\log(x - 1)}{2} + \frac{\log(x^{2} + 1)}{4} + \frac{\tan^{- 1}x}{2} + c

D

None of these

Answer

x(x+2)2+log⁡(x−1)2−log⁡(x2+1)4−tan⁡−1x2+c\frac{x(x + 2)}{2} + \frac{\log(x - 1)}{2} - \frac{\log(x^{2} + 1)}{4} - \frac{\tan^{- 1}x}{2} + c

Explanation

Solution

I2=∫x2+4x+3dx=∫(x+2)2−12dxI _ { 2 } = \int \sqrt { x ^ { 2 } + 4 x + 3 } d x = \int \sqrt { ( x + 2 ) ^ { 2 } - 1 ^ { 2 } } d x =∫x4−1(x−1)(x2+1)dx+∫1(x−1)(x2+1)dx= \int_{}^{}{\frac{x^{4} - 1}{(x - 1)(x^{2} + 1)}dx + \int_{}^{}{\frac{1}{(x - 1)(x^{2} + 1)}dx}} =12(x+2)x2+4x+3−12log⁡∣x+2+x2+4x+3∣+c2= \frac { 1 } { 2 } ( x + 2 ) \sqrt { x ^ { 2 } + 4 x + 3 } - \frac { 1 } { 2 } \log \left| x + 2 + \sqrt { x ^ { 2 } + 4 x + 3 } \right| + c _ { 2 } =∫(x+1)dx+∫dx(x−1)(x2+1)= \int_{}^{}{(x + 1)dx + \int_{}^{}\frac{dx}{(x - 1)(x^{2} + 1)}} =x22+x+[12log⁡(x−1)−14log⁡(x2+1)−12tan⁡−1x]+c= \frac { x ^ { 2 } } { 2 } + x + \left[ \frac { 1 } { 2 } \log ( x - 1 ) - \frac { 1 } { 4 } \log \left( x ^ { 2 } + 1 \right) - \frac { 1 } { 2 } \tan ^ { - 1 } x \right] + c =x(x+2)2+log⁡(x−1)2−log⁡(x2+1)4−tan⁡−1x2+c= \frac { x ( x + 2 ) } { 2 } + \frac { \log ( x - 1 ) } { 2 } - \frac { \log \left( x ^ { 2 } + 1 \right) } { 4 } - \frac { \tan ^ { - 1 } x } { 2 } + c.