Evaluate the following integral: \int \frac {1}{x^2\sqrt{x^2+4}}dx

Question:

Evaluate the following integral: {eq}\int \frac {1}{x^2\sqrt{x^2+4}}dx {/eq}

Indefinite integration:

We are given a rational function and we need to compute the indefinite integral. To solve many integrals, we take help of a trigonometric function, i.e we change the given variable by replacing it with a short-term trigonometric substitute to make the process simple.

Next, we'll simplify the answer by using the identity {eq}sin\left( arctan(x) \right)=\dfrac{x}{\sqrt {1+x^2}} {/eq}.

Answer and Explanation:

{eq}\displaystyle\int \frac {1}{x^2\sqrt{x^2+4}} \ dx {/eq}


Apply Trig Substitution: {eq}x=2 \tan (u) \rightarrow \ dx =2 \sec^2 (u) \ du {/eq}

{eq}=\displaystyle \int \frac{2 \sec^2 (u)}{ 49\tan^2 (u)\sqrt{4\tan^2 (u) + 4}} \ du {/eq}


Apply trig-identity: {eq}\displaystyle \tan^2 (u) + 1 = \sec^2 (u) {/eq}

{eq}=\displaystyle \int \frac{\sec^2 (u)}{ 4\tan^2 (u)\sec (u) } \ du {/eq}

{eq}=\displaystyle \int \frac{\sec (u)}{4 \tan^2 (u) } \ du {/eq}


Apply trig-identity: {eq}\displaystyle \sec (u) = \frac{1}{\cos(u)} \ , \ \tan(u) = \frac{\sin(u)}{\cos(u)} {/eq}

{eq}=\displaystyle \int \frac{ \cos(u)}{4sin^2 (u)} \ du {/eq}


Apply u-substitution: {eq}\displaystyle v = \sin (u) \rightarrow \ dv = \cos(u) \ du {/eq}

{eq}=\displaystyle \int \frac{1}{4v^2} \ dv {/eq}


Take the constant outside the integral:

{eq}=\displaystyle \dfrac{1}{4} \int v^{-2} \ dv {/eq}

{eq}=\displaystyle \dfrac{1}{4} \dfrac{v^{-2+1} }{-2+1}+C {/eq}

{eq}=\displaystyle \dfrac{1}{4} \dfrac{v^{-1} }{-1}+C {/eq}

{eq}=\displaystyle - \dfrac{1}{4v}+C {/eq}


Substitute back {eq}x= 2 \tan (u) \Rightarrow u=arctan(x/7) {/eq} and {eq}v= \sin (u) \Rightarrow v=sin\left( arctan(x/2) \right) {/eq}

{eq}=\displaystyle - \dfrac{1 }{4sin\left( arctan(x/2) \right)}+C {/eq}


Now, {eq}sin\left( arctan(x) \right)=\dfrac{x}{\sqrt {1+x^2}}+C {/eq}

{eq}=\displaystyle- \dfrac{\sqrt{x^2+4}}{4x}+C {/eq}


Therefore solution is:

{eq}\displaystyle {\boxed{ \int \frac{1}{x^{2} \sqrt{x^{2} + 4}}dx=- \dfrac{\sqrt{x^2+4}}{4x}+C}} {/eq}


Learn more about this topic:

Loading...
Indefinite Integral: Definition, Rules & Examples

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14
20K

Related to this Question

Explore our homework questions and answers library