Question

\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)} dx\]

Answer 1

We have,

\[I = \int\frac{2x dx}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)}\]

\[\text{Putting }x^2 = t\]
\[ \Rightarrow 2x\ dx = dt\]
\[ \therefore I = \int\frac{dt}{\left( t + 1 \right) \left( t + 3 \right)}\]
\[\text{Let }\frac{1}{\left( t + 1 \right) \left( t + 3 \right)} = \frac{A}{t + 1} + \frac{B}{t + 3}\]
\[ \Rightarrow \frac{1}{\left( t + 1 \right) \left( t + 3 \right)} = \frac{A \left( t + 3 \right) + B \left( t + 1 \right)}{\left( t + 1 \right) \left( t + 3 \right)}\]
\[ \Rightarrow 1 = A \left( t + 3 \right) + B \left( t + 1 \right)\]
Putting t + 3 = 0
\[ \Rightarrow t = - 3\]
\[1 = A \times 0 + B \left( - 3 + 1 \right)\]
\[ \Rightarrow B = - \frac{1}{2}\]
Putting t + 1 = 0
\[ \Rightarrow t = - 1\]
\[1 = A \left( - 1 + 3 \right) + B \left( - 1 + 1 \right)\]
\[ \Rightarrow 1 = A \times 2 + B \times 0\]
\[ \Rightarrow A = \frac{1}{2}\]
Then,
\[I = \frac{1}{2}\int\frac{dt}{t + 1} - \frac{1}{2}\int\frac{dt}{t + 3}\]
\[ = \frac{1}{2} \log \left| t + 1 \right| - \frac{1}{2} \log \left| t + 3 \right| + C\]
\[ = \frac{1}{2} \log \left| \frac{t + 1}{t + 3} \right| + C\]
\[ = \frac{1}{2} \log \left| \frac{x^2 + 1}{x^2 + 3} \right| + C\]