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x3dx1+x2 is equal to 

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a
131+x22+x2+C
b
131+x2x2−1+C
c
131+x23/2+C
d
131+x2x2−2+C

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detailed solution

Correct option is D

I=∫x3dx1+x2=∫x⋅x2dx1+x2 Let t=1+x2⇒dtdx=x1+x2 Now, I=∫t2−1dt=t33−t+C=t3t2−3+C∴ I=131+x2x2−2+C


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