Question

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

Answer 1

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

\[\text{Let }x^2 = y\]
\[ \Rightarrow 2xdx = dy\]
\[ \Rightarrow dx = \frac{dy}{2x}\]

\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)^2}dx\]
\[ = \int\frac{dy}{\left( y + 1 \right) \left( y + 2 \right)^2}\]
\[\text{Let }\frac{1}{\left( y + 1 \right) \left( y + 2 \right)^2} = \frac{A}{y + 1} + \frac{B}{y + 2} + \frac{C}{\left( y + 2 \right)^2} . . . . . \left( 1 \right)\]
\[ \Rightarrow 1 = A \left( y + 2 \right)^2 + B\left( y + 1 \right)\left( y + 2 \right) + C\left( y + 1 \right) . . . . . \left( 2 \right)\]
\[\text{Putting y = - 2 in (2)}\]
\[1 = C\left( - 2 + 1 \right)\]
\[ \Rightarrow C = - 1\]

\[\text{Putting y = - 1 in (2)}\]
\[1 = A \left( - 1 + 2 \right)^2 \]
\[ \Rightarrow 1 = A\left( 1 \right)\]
\[ \Rightarrow A = 1\]

\[\text{Putting y = 0 in (2)}\]
\[1 = 4A + B\left( 2 \right) + C\]
\[ \Rightarrow 1 = 4 + 2B - 1\]
\[ \Rightarrow 1 = 3 + 2B\]
\[ \Rightarrow - 2 = 2B\]
\[ \Rightarrow B = - 1\]

\[\text{Substituting the values of A, B and C in (1)}\]

\[\frac{1}{\left( y + 1 \right) \left( y + 2 \right)^2} = \frac{1}{y + 1} - \frac{1}{y + 2} - \frac{1}{\left( y + 2 \right)^2}\]
\[ \Rightarrow \int\frac{dy}{\left( y + 1 \right) \left( y + 2 \right)^2} = \int\frac{dy}{y + 1} - \int\frac{dy}{y + 2} - \int\frac{dy}{\left( y + 2 \right)^2}\]
\[ = \log\left| y + 1 \right| - \log\left| y + 2 \right| + \frac{1}{y + 2} + C\]

\[\text{Hence, }\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)^2}dx= \log\left| x^2 + 1 \right| - \log\left| x^2 + 2 \right| + \frac{1}{x^2 + 2} + C\]