If ∫1x4+1dx=A tan−1x2−12x−B logx2−2x+1x2+2x+1+C then

I=∫1x4+1dx =12∫x2+1-x2-1x4+1dx =12∫x2+1x4+1dx-12∫x2-1x4+1dx =12∫1+1x2x2+1x2dx-12∫1-1x2x2+1x2dx =Put x-1x=t Put x+1x=y 1+1x2dx=dt 1-1x2dx=dy Now t2=x-1x2=x2+1x2-2 Now y2=x+1x2=x2+1x2+2 t2+2=x2+1x2 y2-2=x2+1x2 =12∫dtt2+2 -12∫dyy2-2 =1212tan-1t2-12122logy-2y+2+C =122tan-1x-1x2-142logx+1x-2x+1x+2+C =122tan-1x2-12x-142logx2+1-2xx2+1+2x+C ∴A=122, B=142 2B=A

If ∫1x4+1dx=A tan−1x2−12x−B logx2−2x+1x2+2x+1+C then

I=∫1x4+1dx =12∫x2+1-x2-1x4+1dx =12∫x2+1x4+1dx-12∫x2-1x4+1dx =12∫1+1x2x2+1x2dx-12∫1-1x2x2+1x2dx =Put x-1x=t Put x+1x=y 1+1x2dx=dt 1-1x2dx=dy Now t2=x-1x2=x2+1x2-2 Now y2=x+1x2=x2+1x2+2 t2+2=x2+1x2 y2-2=x2+1x2 =12∫dtt2+2 -12∫dyy2-2 =1212tan-1t2-12122logy-2y+2+C =122tan-1x-1x2-142logx+1x-2x+1x+2+C =122tan-1x2-12x-142logx2+1-2xx2+1+2x+C ∴A=122, B=142 2B=A

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If $\int \frac{1}{x^{4} + 1} d x = A \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2 x}}) - B \log |\frac{x^{2} - \sqrt{2 x} + 1}{x^{2} + \sqrt{2 x} + 1}| + C$ then

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$2 A = B$

$A = - 2 B$

$A = 2 B$

$2 A = - B$

answer is A.

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Detailed Solution

$I = \int \frac{1}{x^{4} + 1} d x = \frac{1}{2} \int \frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + 1} d x = \frac{1}{2} \int \frac{x^{2} + 1}{x^{4} + 1} d x - \frac{1}{2} \int \frac{x^{2} - 1}{x^{4} + 1} d x = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x = P u t x - \frac{1}{x} = t P u t x + \frac{1}{x} = y (1 + \frac{1}{x^{2}}) d x = d t (1 - \frac{1}{x^{2}}) d x = d y N o w t^{2} = {(x - \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} - 2 N o w y^{2} = {(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2 t^{2} + 2 = x^{2} + \frac{1}{x^{2}} y^{2} - 2 = x^{2} + \frac{1}{x^{2}} = \frac{1}{2} \int \frac{d t}{t^{2} + 2} - \frac{1}{2} \int \frac{d y}{y^{2} - 2} = \frac{1}{2} \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t}{\sqrt{2}}) - \frac{1}{2} \frac{1}{2 \sqrt{2}} \log |\frac{y - \sqrt{2}}{y + \sqrt{2}}| + C = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{4 \sqrt{2}} \log |\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}| + C = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log |\frac{x^{2} + 1 - \sqrt{2} x}{x^{2} + 1 + \sqrt{2} x}| + C ∴ A = \frac{1}{2 \sqrt{2}}, B = \frac{1}{4 \sqrt{2}} 2 B = A$