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a
231−x232−1−x2525+1−x2+C
b
231−x232−1−x2525−1−x2+C
c
231−x232+1−x2525−1−x2+C
d
None
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Correct option is B
∫x51−x2dx1−x2=t2xdx=−tdtI=∫x22xdx1−x2dx=−∫1−t22ttdt=−∫1+t4−2t2dt=2t33−t55−t+C=231−x232−1−x2525−1−x2+CSimilar Questions
is equal to
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