Integrals — Class 12 Maths Solution

Let $I = \int {\cfrac{{5x\,dx}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}}$

Let $\cfrac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{{Bx + C}}{{{x^2} + 9}}$

$\Rightarrow$ $5x = A\left( {{x^2} + 9} \right) + \left( {Bx + C} \right)\left( {x + 1} \right)$

….(i)
Putting $x = - 1$ in (i),

we get
$5\left( { - 1} \right) = A\left( {1 + 9} \right)$ $\Rightarrow$ $A = - \cfrac{1}{2}$

Comparing coefficients of ${x^2}$ in $(i)$ ,

we get
$0 = A + B$ $\Rightarrow$ $B = \cfrac{1}{2}$

Putting $x = 0$ in (i),

we get$0 = 9A + C$
$\Rightarrow$ $C = - 9A = - 9\left( { - \cfrac{1}{2}} \right) = \cfrac{9}{2}$

$\therefore$ $I = \int {\cfrac{{ - \cfrac{1}{2}}}{{x + 1}}dx} + \int {\cfrac{{\cfrac{1}{2}x + \cfrac{9}{2}}}{{{x^2} + 9}}dx}$

$= - \cfrac{1}{2}\log \left( {x + 1} \right) + \cfrac{1}{4}\int {\cfrac{{2x}}{{{x^2} + 9}}} dx + \cfrac{9}{2}\int {\cfrac{{dx}}{{{x^2} + {3^2}}} + C}$

$= - \cfrac{1}{2}\log \left( {x + 1} \right) + \cfrac{1}{4}\log \left( {{x^2} + 9} \right) + \cfrac{9}{2} \times \cfrac{1}{3}{\tan ^{ - 1}}\cfrac{x}{3} + C$

$= - \cfrac{1}{2}\log \left( {x + 1} \right) + \cfrac{1}{4}\log \left( {{x^2} + 9} \right) + \cfrac{3}{2}{\tan ^{ - 1}}\cfrac{x}{3} + C$