Question

\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]

Answer 1

We have,

\[I = \int\frac{dx}{x \left( x^n + 1 \right)}\]

\[ = \int\frac{x^{n - 1} dx}{x^{n - 1} x \left( x^n + 1 \right)}\]

\[ = \int\frac{x^{n - 1} dx}{x^n \left( x^n + 1 \right)}\]

Putting `x^n = t`

\[ \Rightarrow n x^{n - 1} dx = dt\]

\[ \Rightarrow x^{n - 1} dx = \frac{dt}{n}\]

\[ \therefore I = \frac{1}{n}\int\frac{dt}{t \left( t + 1 \right)}\]

\[\text{Let }\frac{1}{t \left( t + 1 \right)} = \frac{A}{t} + \frac{B}{t + 1}\]

\[ \Rightarrow \frac{1}{t \left( t + 1 \right)} = \frac{A \left( t + 1 \right) + Bt}{t \left( t + 1 \right)}\]

\[ \Rightarrow 1 = A \left( t + 1 \right) + Bt\]

Putting `t + 1 = 0`

\[ \Rightarrow t = - 1\]

\[1 = A \times 0 + B \left( - 1 \right)\]

\[ \Rightarrow B = - 1\]

Putting `t = 0`

\[1 = A \left( 0 + 1 \right) + B \times 0\]

\[ \Rightarrow A = 1\]

Then,

\[I = \frac{1}{n}\int\frac{dt}{t} - \frac{1}{n}\int\frac{dt}{t + 1}\]

\[ = \frac{1}{n} \log \left| t \right| - \frac{1}{n}\log \left| t + 1 \right| + C\]

\[ = \frac{1}{n} \log \left| \frac{t}{t + 1} \right| + C\]

\[ = \frac{1}{n} \log \left| \frac{x^n}{x^n + 1} \right| + C\]