Evaluate the integral:

Question:

Evaluate the integral:

$\int \sqrt{16 x^{2}+25} d x$

Solution:

Key points to solve the problem:

- Such problems require the use of the method of substitution along with a method of integration by parts. By the 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, $I=\int \sqrt{16 x^{2}+25} d x$

We have:

$I=\int \sqrt{16 x^{2}+25} d x=\int \sqrt{(4 x)^{2}+5^{2}} d x$

$\Rightarrow I=\int 4 \sqrt{x^{2}+\left(\frac{5}{4}\right)^{2}} d x$

As I match with the form:

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

$\therefore I=4\left\{\frac{x}{2} \sqrt{x^{2}+\left(\frac{5}{4}\right)^{2}}+\frac{\frac{25}{16}}{2} \log \left|x+\sqrt{x^{2}+\left(\frac{5}{4}\right)^{2}}\right|\right\}$

$\Rightarrow I=\frac{x}{2} \sqrt{16 x^{2}+25}+\frac{25}{8} \log \left|x+\sqrt{x^{2}+\frac{25}{16}}\right|+C$