The integral ∫dx/(x^2x4+1^34) equals to

Correct answer: (A) -x4+1x4^14+c. I=∫dx/(x^2x4+1^34) =∫dx/(x^51+1x4^34) Put, 1+(1x^4=t⇒-4x^5dx=dt∴ I=1)/4∫(-dtt^34 = - 1)/4 × ∫ t^(- 3 4) dt =(-1)/4t^141/4+c =-1+1x4^14+c, where c is the constant of integration.

Q37 / 90 · JEE Main 2015 (04 Apr) · Mathematics

The integral $\int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$ equals to

Correct answer
A
$- {(\frac{x^{4} + 1}{x^{4}})}^{\frac{1}{4}} + c$
B
${(\frac{x^{4} + 1}{x^{4}})}^{\frac{1}{4}} + c$
C
${(x^{4} + 1)}^{\frac{1}{4}} + c$
D
$- {(x^{4} + 1)}^{\frac{1}{4}} + c$

Solution

I = \int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}

= \int \frac{d x}{x^{5} {(1 + \frac{1}{x^{4}})}^{\frac{3}{4}}}

Put,

1 + \frac{1}{x^{4}} = t

\Rightarrow \frac{- 4}{x^{5}} d x = d t

∴ I = \frac{1}{4} \int \frac{- d t}{t^{\frac{3}{4}}}

= - \frac{1}{4} \times \int t^{\frac{- 3}{4}} dt

= \frac{- 1}{4} (\frac{t^{\frac{1}{4}}}{\frac{1}{4}}) + c

= - {(1 + \frac{1}{x^{4}})}^{\frac{1}{4}} + c

, where

c

is the constant of integration.

Hence, the correct answer is option: 'A'.

The integral ∫dxx2x4+134 equals to

Solution

Relevant Questions

The integral $\int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$ equals to