Question
Mathematics Question on Integrals of Some Particular Functions
If ∫x3(1+x6)32dx=f(x)(1+x6)31+C, where C is a constant of integration, then the function f(x) is equal to-
A
−6x31
B
x23
C
−2x21
D
−2x31
Answer
−2x31
Explanation
Solution
∫x3(1+x6)32dx=xf(x)(1+x6)31+c
∫x7(x61+1)32dx=xf(x)(1+x6)31+c
Let t=x61+1
dt=x7−6dx
=−61∫t32dt=−21t31
=−21(x61+1)31=−21x2(1+x6)31
∴f(x)=−2x31