Skip to content
Miscellaneous Exercise · Q4

Q.Integrate the function 1x2(x4+1)3/4\frac{1}{x^2(x^4+1)^{3/4}}

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
Appeared in past exams:AP EAPCET 2026· Set eng-2026-05-13-AN· 1mexactKEAM 2024· Set eng-2024-0609· 4mexactMHT-CET 2024· Set pcm-2024-05-10-E· 2mexactMHT-CET 2023· Set pcm-2023-05-10-E· 2mexactCOMEDK 2021· Set 2021· 1mexact
71% · 266/373 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Taking x4x^4 out of the bracket and substituting t=1+x−4t=1+x^{-4} reduces the integral to −14∫t−3/4 dt-\tfrac14\int t^{-3/4}\,dt, giving −(1+1x4)1/4+C-\left(1+\dfrac{1}{x^4}\right)^{1/4}+C.

We integrate ∫dxx2 (x4+1)3/4\displaystyle\int\frac{dx}{x^2\,(x^4+1)^{3/4}}.

1. Pull x4x^4 out of the bracket. Since (x4+1)3/4=x3 ⁣(1+x−4)3/4(x^4+1)^{3/4}=x^3\!\left(1+x^{-4}\right)^{3/4},

1x2 (x4+1)3/4=1x5(1+x−4)3/4.\frac{1}{x^2\,(x^4+1)^{3/4}}=\frac{1}{x^5\left(1+x^{-4}\right)^{3/4}}.

2. Substitute t=1+x−4t=1+x^{-4}, so dt=−4x−5 dx⇒dxx5=−14 dtdt=-4x^{-5}\,dx\Rightarrow \dfrac{dx}{x^5}=-\dfrac14\,dt:

∫t−3/4(−14)dt=−14∫t−3/4 dt.\int t^{-3/4}\left(-\frac14\right)dt=-\frac14\int t^{-3/4}\,dt. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.