I=∫x2(x4+1)43dx
=∫x5(1+x41)43dx
Put, 1+x41=t⇒x5−4dx=dt
∴I=41∫t43−dt
=−41×∫t4−3dt=4−1(41t41)+c
=−(1+x41)41+c, where c is the constant of integration.
JEE Main 2015 — Mathematics Calculus
The integral ∫x2(x4+1)43dx equals to
Held on 4 Apr 2015 · Verified 6 Jul 2026.
−(x4x4+1)41+c
(x4x4+1)41+c
(x4+1)41+c
−(x4+1)41+c
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