# Evaluate the integral. {eq}\displaystyle \int_1^5 \dfrac {dx} {x^2 \sqrt {4 + x^2}} {/eq}

## Question:

Evaluate the integral.

{eq}\displaystyle \int_1^5 \dfrac {dx} {x^2 \sqrt {4 + x^2}} {/eq}

## Definite Integrals:

Although there is no specific method to solve the definite or indefinite integrals, but when the integrand function is an expression of the following forms: {eq}\displaystyle \sqrt{a^2+u^2} {/eq}, {eq}\displaystyle \sqrt{u^2-a^2} {/eq}, {eq}\displaystyle \sqrt{a^2-u^2} {/eq}. In these cases, we can use the method of trigonometric substitution, in addition to the fundamental theorem of calculus.

.

We can rewrite the integral:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{dx}{x^2\sqrt{2^2+x^2}}\\ {/eq}

Here, how the integrand is of the form {eq}\displaystyle \sqrt{a^2+u^2} {/eq}, we use the trigonometric substitution:

{eq}\displaystyle\,x=2\tan\theta\Rightarrow{\tan\theta=\frac{x}{2}} \\ \displaystyle\,dx=2\sec^2\theta\,d\theta \\ \displaystyle \sqrt{2^2+x^2}=\sqrt{2^2+ 2^2\tan^2\theta} \\ \displaystyle \sqrt{2^2+x^2}=\sqrt{2^2(1+ \tan^2\theta)} \\ \displaystyle \sqrt{2^2+x^2}=\sqrt{4}\cdot{\sqrt{1+\tan^2\theta}} \\ \displaystyle \sqrt{2^2+x^2}=2\cdot{\sqrt{1+\tan^2\theta}} \\ {/eq}

But we know by trigonometric identity that:

{eq}\displaystyle 1+\tan^2\theta=\sec^2\theta \\ {/eq}

Then:

{eq}\displaystyle \sqrt{2^2+x^2}=2\cdot{\sqrt{\sec^2\theta}} \\ \displaystyle \sqrt{2^2+x^2}=2\sec\theta \\ {/eq}

By replacing in the original integral:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{dx}{x^2\sqrt{2^2+x^2}}\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{2\sec^2\theta}{4\tan^2\,2\sec\theta}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{2\sec^2\theta}{4\tan^2\theta\cdot{2\sec\theta}}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\sec\theta}{\tan^2\theta}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\frac{1}{\cos\theta}}{\frac{\sin^2\theta}{\cos^2\theta}}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\cos^2\theta}{\cos\theta\sin\theta}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\cos\theta}{\sin^2\theta}d\theta\\ {/eq}

Doing a change of variable:

{eq}\displaystyle\,w= \sin\theta\\ \displaystyle\,dw=\cos\theta\,d\theta\\ {/eq}

Substituting in the integral:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{dw}{w^2}\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5 w^{-2}dw\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{1}{w}\right)\bigg|_{1}^{5}\\ {/eq}

But:

{eq}\displaystyle\,w=\sin\theta\\ {/eq}

And

{eq}\displaystyle\sin\theta=\frac{x}{\sqrt{4+x^2}}\\ {/eq}

We take the original integral and substitute:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{1}{\frac{x}{\sqrt{4+x^2}}}\right)\bigg|_{1}^{5}\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{\sqrt{4+x^2}}{x}\right)\bigg|_{1}^{5}\\ {/eq}

Applying the fundamental theorem of calculus:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{\sqrt{4+(5)2}}{5}- \frac{\sqrt{4+(1)^2}}{(1)}\right)\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{\sqrt{29}}{5}-\sqrt{5}\right)\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\boxed{\frac{1}{4}\sqrt{5}-\frac{\sqrt{29}}{20}} {/eq}