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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Integrate the function $\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$ 
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Answer

Let $I=\frac{1}{x^{2} \cdot\left(x^{4}+1\right)^{\frac{3}{4}}}$ 
Taking $x^{4}$ common from the denominator, we get
$I =\int \frac{1 d x}{x^{2}\left(x^{4}\right)^{\frac{3}{4}}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$
$=\int \frac{d x}{x^{2}\left(x^{3}\right)\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$
$=\int \frac{d x}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$
$\text { Let } t=1+\frac{1}{x^{4}} \Rightarrow-\frac{d t}{4}=\frac{d x}{x^{5}}$
$\Rightarrow \int \frac{1}{x^{2} \cdot\left(x^{4}+1\right)^{\frac{3}{4}}} \cdot d x$  $=\frac{-1}{4} \int \frac{d t}{t^{\frac{3}{4}}}$ 
= $\frac{-1}{4}\left(\frac{t^{\frac{1}{4}}}{\frac{1}{4}}\right)+C$ 
= $-t^{\frac{1}{4}}+c$
=$-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$

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