Question

Evaluate:$\int \frac{1}{x\sqrt{a^n+x^n}}dx$

• A. $\log \frac{\sqrt{a^n+x^n}-\sqrt{a^n}}{2}+c$
• B. $\frac{1}{n\sqrt{a^n}}\log \left[\frac{\sqrt{a^n+x^n}-\sqrt{a^n}}{\sqrt{a^n+x^n}+\sqrt{a^n}}\right]+c$
• C. $\log \left[\frac{\sqrt{a^n+x^n}+\sqrt{a^n}}{\sqrt{a^n+x^n}-\sqrt{a^n}}\right]+c$
• D. $\frac{1}{n\sqrt{a^n}}\log \left[\frac{\sqrt{a^n+x^n}+\sqrt{a^n}}{\sqrt{a^n+x^n}-\sqrt{a^n}}\right]+c$

Answer 1

The correct option is B $\frac{1}{n\sqrt{a^n}}\log \left[\frac{\sqrt{a^n+x^n}-\sqrt{a^n}}{\sqrt{a^n+x^n}+\sqrt{a^n}}\right]+c$

Multiply, denominator and numerator by $x^{n-1}$

$I=\int \frac{1}{x\sqrt{a^n+x^n}}dx=\int \frac{x^{n-1}}{x^n\sqrt{a^n+x^n}}dx\frac{x^n}{a^n}=t\left(substitution\right)\therefore n{x}^{n-1}dx=a^{n}dtI=\int \frac{x^{n-1}}{a^{n}t\sqrt{a^n+a^{n}t}}\frac{a^{n}dt}{n{x}^{n-1}}=\frac{1}{n\sqrt{a^n}}\int \frac{dt}{t\sqrt{1+t}}t={\tan }^{2}ydt=2\tan y{\sec }^{2}ydysubstitute,I=\frac{1}{n\sqrt{a^n}}\int \frac{2\tan y{\sec }^{2}ydy}{{\tan }^{2}y\sec y}=\frac{2}{n\sqrt{a^n}}\int \csc ydyI=\frac{2}{n\sqrt{a^n}}[-\log (\csc y+\cot y)]I=\frac{1}{n\sqrt{a^n}}\log \left(\frac{{\sin }^{2}y}{{(1+\cos y)}^2}\right)+CI=\frac{1}{n\sqrt{a^n}}\log \left(\frac{{\sin }^{2}y}{{{\cos }^{2}y(\sec y+1)}^2}\right)+CI=\frac{1}{n\sqrt{a^n}}\log \left(\frac{{\tan }^{2}y}{{(\sqrt{1+{\tan }^{2}y}+1)}^2}\right)+CI=\frac{1}{n\sqrt{a^n}}\log \left(\frac{t}{{(\sqrt{1+t}+1)}^2}\right)+CI=\frac{1}{n\sqrt{a^n}}\log \left(\frac{(\sqrt{1+t}-1)(\sqrt{1+t}+1)}{{(\sqrt{1+t}+1)}^2}\right)+CI=\frac{1}{n\sqrt{a^n}}\log \left(\frac{(\sqrt{1+t}-1)}{(\sqrt{1+t}+1)}\right)+Cresubstitute,I=\frac{1}{n\sqrt{a^n}}\log \left[\frac{\sqrt{a^n+x^n}-a^n}{\sqrt{a^n+x^n}+a^n}\right]+C$