The integral $\int \frac{\mathrm{dx}}{\mathrm{x}^{2}\left(\mathrm{x}^{4}+1\right)^{3 / 4}}$ equals :

(C) $-\left(\frac{x^{4}+1}{x^{4}}\right)^{1 / 4}+c$

(A) $\left(x^{4}+1\right)^{1 / 4}+c$

(B) $-\left(x^{4}+1\right)^{1 / 4}+c$

(D) $\left(\frac{x^{4}+1}{x^{4}}\right)^{1 / 4}+c$

JEE Main · 2015 · 04 Apr 2015 · Offline · Indefinite Integrals

The integral

\int \frac{dx}{x ^{2} ( x ^{4} + 1 ) ^{3/4}}

equals :

Answer: (C)

Pull a factor of

x^{4}

out of the awkward root:

(x^{4} + 1)^{3/4} = x^{3} (1 + x^{- 4})^{3/4}

, so the integrand becomes

\frac{1}{x ^{5} ( 1 + x ^{- 4} ) ^{3/4}}

. The substitution

t = 1 + x^{- 4}

, with

d t = - 4 x^{- 5} d x

, converts the integral into

- \frac{1}{4} \int t^{- 3/4} d t = - t^{1/4} + c

. Restoring

t

, this is

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

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option