#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 71

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

Hint: Use substitution method to solve this integral

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

Solution:

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

$=\int \frac{\frac{1}{x^{3}}}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}} \cdot \frac{1}{x^{3}}} d x \quad\left[\text { Dividing numerator and denominator by } \frac{1}{x^{3}}\right]$

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

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

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

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

\begin{aligned} &\text { Put } \frac{1}{x^{4}}=\mathrm{t} \Rightarrow \frac{-4 d x}{x^{5}}=d t \\ &\Rightarrow d x=\frac{x^{5}}{-4} d t \text { then } \end{aligned}

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

\begin{aligned} &\text { Again put } 1+t=u \Rightarrow d t=d u \text { then }\\ &I=\frac{-1}{4} \int(u)^{\frac{-3}{4}} d u=\frac{-1}{4} \frac{u^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

$=\frac{-1}{4} \frac{u^{\frac{1}{4}}}{\frac{1}{4}}+c=\frac{-1}{4} \times 4 \cdot u^{\frac{1}{4}}+c$

$=-u^{\frac{1}{4}}+c=-(1+t)^{\frac{1}{4}}+c \quad[\because u=1+t]$

$=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c \quad\left[\because t=\frac{1}{x^{4}}\right]$