Question

\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

Answer 1

\[I=\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

Using partial fraction,

\[\frac{x}{(1 + x)(1 + x^2 )}\frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}\]

\[x = A(1 + x^2 ) + (Bx + C)(1 + x)\]

\[x = A + A x^2 + Bx + B x^2 + C + Cx\]

\[B + C = 1\]

\[A + C = 0\]

\[A + B = 0\]

\[so, A = \frac{- 1}{2}, B = \frac{1}{2}, C = \frac{1}{2}\]

Putting the values of A, B and C we get

\[\frac{\frac{- 1}{2}}{1 + x} + \frac{\frac{1}{2}x + \frac{1}{2}}{1 + x^2}\]

\[ = \frac{- 1}{2}\left[ \frac{1}{1 + x} \right] + \frac{1}{2}\left[ \frac{x + 1}{1 + x^2} \right]\]

\[\text{Therefore, }I = \int_0^\infty \frac{- 1}{2}\left[ \frac{1}{1 + x} \right] + \frac{1}{2}\left[ \frac{x + 1}{1 + x^2} \right]\]

\[I = \frac{- 1}{2} \left[ \log\left| 1 + x \right| \right]_0^\infty + \frac{1}{2} \int_0^\infty \left[ \frac{x}{1 + x^2} + \frac{1}{1 + x^2} \right]\]

\[I = \frac{- 1}{2} \left[ log\left| 1 + x \right| \right]_0^\infty + \frac{1}{2 \times 2} \int_0^\infty \left[ \frac{2x}{1 + x^2} \right] + \frac{1}{2} \int_0^\infty \frac{1}{1 + x^2}\]

\[I = \frac{- 1}{2} \left[ \log\left| 1 + x \right| \right]_0^\infty + \frac{1}{4} \left[ \log\left| 1 + x^2 \right| \right]_0^\infty + \left[ \frac{1}{2}ta n^{- 1} x \right]_0^\infty \]

\[I = \frac{- 1}{2} \left[ log\left| 1 + x \right| \right]_0^\infty + \frac{1}{2} \times \frac{1}{2} \left[ log\left| 1 + x^2 \right| \right]_0^\infty + \left[ \frac{1}{2}ta n^{- 1} x \right]_0^\infty \]

\[I = \frac{1}{2} \left[ \log\frac{\sqrt{x^2 + 1}}{x + 1} \right]_0^\infty + \left[ \frac{1}{2}ta n^{- 1} x \right]_0^\infty \]

\[I = \frac{1}{2} \left[ log\frac{\sqrt{1 + \frac{1}{x^2}}}{1 + \frac{1}{x}} \right]_0^\infty + \left[ \frac{1}{2}ta n^{- 1} x \right]_0^\infty \]

\[I = \frac{1}{2}\left[ 0 \right] + \frac{1}{2}\left[ ta n^{- 1} \infty - ta n^{- 1} 0 \right]\]

`I=pi/4`