Question

\[\int\frac{1}{x \left( x^4 + 1 \right)} dx\]

Answer 1

We have,
\[I = \int \frac{dx}{x\left( x^4 + 1 \right)}\]
\[ = \int\frac{x^3 dx}{x^4 \left( x^4 + 1 \right)}\]
\[\text{Putting} x^4 = t\]
\[ \Rightarrow 4 x^3 dx = dt\]
\[ \Rightarrow x^3 dx = \frac{dt}{4}\]
\[ \therefore I = \frac{1}{4}\int\frac{dt}{t\left( t + 1 \right)}\]
\[\text{Let }\frac{1}{t\left( t + 1 \right)} = \frac{A}{t} + \frac{B}{t + 1}\]
\[ \Rightarrow \frac{1}{t\left( t + 1 \right)} = \frac{A\left( t + 1 \right) + Bt}{t\left( t + 1 \right)}\]
\[ \Rightarrow 1 = A\left( t + 1 \right) + Bt\]
\[\text{Putting }t + 1 = 0\]
\[ \Rightarrow t = - 1\]
\[ \therefore 1 = A \times 0 + B\left( - 1 \right)\]
\[ \Rightarrow B = - 1\]
\[\text{Putting }t = 0\]
\[ \therefore 1 = A\left( 1 \right) + B \times 0\]
\[ \Rightarrow A = 1\]
\[ \therefore I = \frac{1}{4}\int\frac{dt}{t} - \frac{1}{4}\int\frac{dt}{t + 1}\]
\[ = \frac{1}{4}\log \left| t \right| - \frac{1}{4}\log \left| t + 1 \right| + C\]
\[ = \frac{1}{4}\log \left| \frac{t}{t + 1} \right| + C\]