Integrals — Maths STD 12 Science — Question

Putting ${x^2} = t \Rightarrow 2xdx = dt$

$\therefore$ From eq. (i),

$I = \int {\frac{{dt}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}} $

$= \frac{1}{2}\int {\frac{2}{{\left( {t + 1} \right)\left( {t + 3} \right)}}dt} $

$= \frac{1}{2}\int {\frac{{\left( {t + 3} \right) - \left( {t + 1} \right)}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}dt} $

$= \frac{1}{2}\int {\left( {\frac{{\left( {t + 3} \right)}}{{\left( {t + 1} \right)\left( {t + 3} \right)}} - \frac{{\left( {t + 1} \right)}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}} \right)dt} $

$= \frac{1}{2}\int {\left( {\frac{1}{{\left( {t + 1} \right)}} - \frac{1}{{\left( {t + 3} \right)}}} \right)dt} $

$= \frac{1}{2}\left[ {\log \left| {t + 1} \right| - \log \left| {t + 3} \right|} \right] + c$

$= \frac{1}{2}\left[ {\log \left| {\frac{{t + 1}}{{t + 3}}} \right|} \right] + c$

$= \frac{1}{2}\log \left| {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + c$

$= \frac{1}{2}\log \left( {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right) + c$.

Which is the required solution.