Integrals — Class 12 Maths Solution

Let $I = \int\limits_1^2 {\cfrac{{5{x^2}}}{{{x^2} + 4x + 3}}} dx$

Since the degree of numerator denominator and is same so the, fraction is improper.

To mark it proper, we have to divide $5{x^2}$ by ${x^2} + 4x + 3$ .

${x^2} + 4x + 3\mathop{\left){\vphantom{1\begin{array}{l}5{x^2}\\5{x^2} + 20x + 15\\\underline { - \,\,\,\,\,\, - \,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,} \\\,\,\,\,\,\,\, - 20x - 15\end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}5{x^2}\\5{x^2} + 20x + 15\\\underline { - \,\,\,\,\,\, - \,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,} \\\,\,\,\,\,\,\, - 20x - 15\end{array}}}} \limits^{\displaystyle\,\,\, 5}$

$\therefore$ $I = \int\limits_1^2 {\left( {5 + \cfrac{{ - 20x - 15}}{{{x^2} + 4x + 3}}} \right)} dx = \int\limits_1^2 {\left( {5 - \cfrac{{20x + 15}}{{{x^2} + 4x + 3}}} \right)} dx$

$\cfrac{{20x + 15}}{{{x^2} + 4x + 3}} = \cfrac{{20x + 15}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{x + 3}}$

$\Rightarrow$ $20x + 15 = A\left( {x + 3} \right) + B\left( {x + 1} \right)$

….(i)
Putting $x = - 1$ in $(i)$,

we get
$- 20 + 15 = A\left( { - 1 + 3} \right)$ $\Rightarrow$ $- 5 = 2A$ $\Rightarrow$ $A = \cfrac{{ - 5}}{2}$

Putting $x = - 3$ in $(i)$,

we get
$- 60 + 15 = B\left( { - 3 + 1} \right)$ $\Rightarrow$ $- 45 = - 2B$ $\Rightarrow$ $B = \cfrac{{45}}{2}$

$\therefore$ $I = \int\limits_1^2 {\left( {5 + \cfrac{5}{{2\left( {x + 1} \right)}} - \cfrac{{45}}{{2\left( {x + 3} \right)}}} \right)dx}$

$= \left[ {5x + \cfrac{5}{2}\log \left( {x + 1} \right) - \cfrac{{45}}{2}\log \left( {x + 3} \right)} \right]_1^2$

$= 5\left( {2 - 1} \right) + \cfrac{5}{2}\left[ {\log 3 - \log 2} \right] - \cfrac{{45}}{2}\left[ {\log 5 - log4} \right]$

$= 5 + \cfrac{5}{2}\log \cfrac{3}{2} - \cfrac{{45}}{2}\log \cfrac{5}{4} = 5 - \cfrac{5}{2}\left( {9\log \cfrac{5}{4} - \log \cfrac{3}{2}} \right)$