# Evaluate the integral:

Question:

Evaluate the integral:

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

Solution:

Key points to solve the problem:

- Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have $\int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{dx}-\int \mathrm{f}^{\prime}(\mathrm{x})\left(\int \mathrm{g}(\mathrm{x}) \mathrm{dx}\right) \mathrm{dx}$

- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:

$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$

$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$

$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

Let, $I=\int x \sqrt{x^{4}+1} d x=\int x \sqrt{\left(x^{2}\right)^{2}+1} d x$

Let, $x^{2}=t$

Differentiating both sides:

$\Rightarrow 2 x d x=d t \Rightarrow x d x=1 / 2 d t$

Substituting $x^{2}$ with $t$, we have:

We have:

$I=\frac{1}{2} \int \sqrt{t^{2}+1} d t=\frac{1}{2} \int \sqrt{t^{2}+1^{2}} d t$

As I match with the form:

$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

$\therefore I$ $=\frac{1}{2}\left\{\frac{\mathrm{t}}{2} \sqrt{\mathrm{t}^{2}+1}+\frac{1}{2} \log \left|\mathrm{t}+\sqrt{\mathrm{t}^{2}+1}\right|\right\}+\mathrm{C}$

$\Rightarrow I$ $=\frac{t}{4} \sqrt{t^{2}+1}+\frac{1}{4} \log \left|t+\sqrt{t^{2}+1}\right|+C$

Putting the value of $\mathrm{t}$ back:

$\Rightarrow I$ $=\frac{x^{2}}{4} \sqrt{\left(x^{2}\right)^{2}+1}+\frac{1}{4} \log \left|x^{2}+\sqrt{\left(x^{2}\right)^{2}+1}\right|+C$

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