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The value of integral dxx1x3 is given equal to

 

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a
13log⁡1−x3+11−x3−1+C
b
13log⁡1−x3−11−x2+1+C
c
23log⁡11−x3+C
d
13log⁡1−x3+C

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detailed solution

Correct option is B

Put 1−x3=t2. Then −3x2dx=2tdt and the entegral becomes −13∫−3x2dxx31−x3=−13∫2tdt1−t2t=23∫dtt2−1=2312log⁡t−1t+1+C=13log⁡1−x3−11−x3+1+C.


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