_ ^ 1 x^2 ( x^4 + 1) ^ 3/4 dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{{x^2}{{({x^4} + 1)}^{3/4}}}}dx = } $

A
$\frac{{{{({x^4} + 1)}^{1/4}}}}{x} + c$
✓
$ - \frac{{{{({x^4} + 1)}^{1/4}}}}{x} + c$
C
$\frac{3}{4}\frac{{{{({x^4} + 1)}^{3/4}}}}{x} + c$
D
$\frac{4}{3}\frac{{{{({x^4} + 1)}^{3/4}}}}{x} + c$

✓

Answer

Correct option: B.

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

b
(b) $\int_{}^{} {\frac{1}{{{x^2}{{({x^4} + 1)}^{3/4}}}}\,dx = \int_{}^{} {\frac{{dx}}{{{x^5}{{\left( {1 + \frac{1}{{{x^4}}}} \right)}^{3/4}}}}} } $
Put $1 + \frac{1}{{{x^4}}} = t \Rightarrow \frac{{ - 4}}{{{x^5}}}\,dx = dt,$ then it reduces to
$ - \frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^{3/4}}}}} = - \frac{1}{4}\frac{4}{1}{t^{1/4}} + c = - {t^{1/4}} + c$
$ = - {\left( {1 + \frac{1}{{{x^4}}}} \right)^{1/4}} + c = - \frac{{{{({x^4} + 1)}^{1/4}}}}{x} + c.$

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