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The value of the integral

1/31/3x41x4cos12x1+x2dx is

 

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a
0
b
π3+log⁡3+13−1
c
13+π2log⁡3+13−1
d
none of these

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detailed solution

Correct option is D

∫−1/31/3 x41−x4cos−1⁡2x1+x2dx=∫−1/31/3 x41−x4π2−sin−1⁡2x1+x2dx=π2∫−1/31/3 x41−x4dxsin−1⁡(−x)=−sin−1⁡x so the last integral is zero)=π∫01/3 −1+11−x4dx=π∫01/3 −1+1211−x2+11+x2dx=−π3+π212log⁡1+x1−x+tan−1⁡x01/3=−π3+π212log⁡3+13−1+π6


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