#### Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 12

$I=\frac{x^{2}}{4} \sqrt{x^{4}+1}+\frac{1}{4} \log \left|x+x^{2} \sqrt{x^{4}+1}\right|+c$

Hints:-

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}

Given:-

$I=\int x \sqrt{x^{4}+1} d x$

Solution:-

Let,
$I=\int x \sqrt{x^{4}+1} d x$

Let, $x^{2}= t$

Differentiating both sides,

\begin{aligned} &\Rightarrow 2 x d x=d t \\\\ &\Rightarrow x d x=\frac{1}{2} d t \end{aligned}

Substituting $x^{2}$ with t, we have

\begin{aligned} &I=\frac{1}{2} \int \sqrt{t^{2}+1} d t \\\\ &I=\frac{1}{2} \int \sqrt{t^{2}+1^{2}} d t \end{aligned}

As I match with the form

\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &I=\frac{1}{2}\left\{\frac{t}{2} \sqrt{t^{2}+1}+\frac{1}{2} \log \left|t+\sqrt{t^{2}+1}\right|\right\}+c \\\\ &I=\frac{t}{4} \sqrt{t^{2}+1}+\frac{1}{4} \log \left|t+\sqrt{t^{2}+1}\right|+c \end{aligned}

Putting the value of t back.

\begin{aligned} &I=\frac{x^{2}}{4} \sqrt{\left(x^{2}\right)^{2}+1}+\frac{1}{4} \log \left|x^{2}+\sqrt{\left(x^{2}\right)^{2}+1}\right|+c \\\\ &I=\frac{x^{2}}{4} \sqrt{x^{4}+1}+\frac{1}{4} \log \left|x+x^{2} \sqrt{x^{4}+1}\right|+c \end{aligned}