Question

\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]

Answer 1

\[\text{We have}, \]
\[I = \int\frac{dx}{x \sqrt{1 + x^n}}\]
\[ = \int\frac{x^{n - 1} dx}{x^{n - 1}\text{  x}^1 \sqrt{1 + x^n}}\]
\[ = \int\frac{x^{n - 1} dx}{x^n \sqrt{1 + x^n}}\]
\[\text{Putting  x}^n = t\]
\[ \Rightarrow \text{ n  x}^{n - 1} dx = dt\]
\[ \Rightarrow x^{n - 1} \text{ dx} = \frac{dt}{n}\]
\[ \therefore I = \frac{1}{n}\int\frac{dt}{t \sqrt{1 + t}}\]
\[\text{ let 1 + t = p}^2 \]
\[ \Rightarrow \text{ dt = 2p dp }\]
\[ \therefore I = \frac{1}{n}\int\frac{\text{ 2p dp}}{\left( p^2 - 1 \right) p}\]
\[ = \frac{2}{n}\int\frac{dp}{p^2 - 1^2}\]
\[ = \frac{2}{n} \times \frac{1}{2} \text{ log} \left| \frac{p - 1}{p + 1} \right| + C\]
\[ = \frac{1}{n} \text{ log} \left| \frac{\sqrt{1 + t} - 1}{\sqrt{1 + t} + 1} \right| + C\]
\[ = \frac{1}{n} \text{ log } \left| \frac{\sqrt{1 + x^n} - 1}{\sqrt{1 + x^n} + 1} \right| + C\]