We have, ∫x3(1+x6)32dx=xf(x)(1+x6)31+C
Now, I=∫x7(x61+1)32dx
Put, t=x61+1⇒dt=−x76dx
∴I=−61∫t32dt=−21t31+C, where C is the constant of integration=−21(x61+1)31+C=−21x2(1+x6)31+C
∴f(x)=−2x31
If ∫x3(1+x6)32dx=xf(x)(1+x6)31+C, where C is a constant of integration, then the function f(x) is equal to
Held on 8 Apr 2019 · Verified 6 Jul 2026.
x23
−2x31
−6x31
−2x21
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