Integrals — Class 12 Maths Solution

.: Let $I = \int {\cfrac{{x\,dx}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}}$

Let $\cfrac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{x + 2}}$

$\Rightarrow$ $x = A\left( {x + 2} \right) + B\left( {x + 1} \right)$

…(i)
Putting $x = - 1$ in (i),

we get $- 1 = A\left( { - 1 + 2} \right)$ $\Rightarrow$ $A = - 1$

Putting $x = - 2$ in (i),

we get $- 2 = B\left( { - 2 + 1} \right)$ $\Rightarrow$ $B = 2$

$\therefore$ $\cfrac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \cfrac{{ - 1}}{{x + 1}} + \cfrac{2}{{x + 2}}$

$\Rightarrow$
$I = \int {\cfrac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}dx = } \int {\left[ {\cfrac{{ - 1}}{{x + 1}} + \cfrac{2}{{x + 2}}} \right]} dx$

$= \int {\cfrac{{ - 1}}{{\left( {x + 1} \right)}}dx + } \int {\cfrac{2}{{x + 2}}} dx$

$= - \log \left| {x + 1} \right| + 2\log \left| {x + 2} \right| + C$

$= - \log \left| {x + 1} \right| + \log {\left( {x + 2} \right)^2} + C = \log \left[ {\cfrac{{{{\left( {x + 2} \right)}^2}}}{{\left| {x + 1} \right|}}} \right] + C$