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

MCQ

$\smallint \frac{{dx}}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}} = $

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

✓

Answer

Correct option: A.

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

a
$\int \frac{1 d x}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$

Taking $x^{4}$ common from denominator

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

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

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

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

$\frac{{dt}}{{dx}} = - \frac{4}{{{x^5}}}$

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

Substituting value of $x$ and $d x$

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

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

${=\frac{-1}{4}\left[\frac{t^{\frac{-3}{4}}+1}{\frac{-3}{4}+1}\right]+C} $

${=\frac{-1}{4}\left(\frac{t^{\frac{1}{4}}}{\frac{1}{4}}\right)+C}$

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

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

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