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JEE Main 20154 April

Exam: JEE Main
Year: 2015
Date: 4 April
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2015 · 4 April · Mathematics · Indefinite Integration

The integral ∫ dx x ^ 2 ( x ^ 4 + 1)(3)/(4) equals to

The integral $∫ \dfrac{dx}{{x}^{2}\left({x}^{4} + 1\right)\dfrac{3}{4}}$ equals to

Options

A $- \left(\dfrac{{x}^{4} + 1}{{x}^{4}}\right)\dfrac{1}{4} + c$
B $\left(\dfrac{{x}^{4} + 1}{{x}^{4}}\right)\dfrac{1}{4} + c$
C $\left({x}^{4} + 1\right)\dfrac{1}{4} + c$
D $- \left({x}^{4} + 1\right)\dfrac{1}{4} + c$

Correct answer

A. $- \left(\dfrac{{x}^{4} + 1}{{x}^{4}}\right)\dfrac{1}{4} + c$

Step-by-step solution

I = ∫ \dfrac{dx}{x^{2}\left(x^{4} + 1\right)\dfrac{3}{4}}

~

= ∫ \dfrac{dx}{x^{5}\left(1 + \dfrac{1}{x4}\right)\dfrac{3}{4}}

Put,~ &

1 + \dfrac{1}{x4} = t⇒ \dfrac{- 4}{x5}dx = dt

∴ I = \dfrac{1}{4} ∫ \dfrac{- dt}{t\dfrac{3}{4}}

= - \dfrac{1}{4} \times ∫ \text{t} \dfrac{- 3}{4} \text{dt}= \dfrac{- 1}{4}\left(\dfrac{t\dfrac{1}{4}}{14}\right) + c

= - \left(1 + \dfrac{1}{x4}\right)\dfrac{1}{4} + c

, where~

c

& is the constant of integration.

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