Let $I=\int\limits_{1}^{2} \frac{5x^2}{x^2+4x+3} dx$ Since the degree of numerator and denominator is same.∴The fraction is improper. To make it proper we have to divide $5x^2$ by $x^2 + 4x + 3.$ $\Rightarrow \frac{5x^2}{x^2+4x+3} = 5 + \frac{-20x-15}{x^2+4x+3}$$I=\int\limits_{1}^{2} \left(5 + \frac{-20x-15}{x^2+4x+3}\right) dx$Let $\frac{20x+15}{x^2+4x+3}$ $= \frac{20x+15}{(x+1)(x+3)}$ $= \frac{A}{x+1} + \frac{B}{x+3}$$\Rightarrow 20x + 15 \equiv A(x+3) + B(x+1)$ ...(i)Putting x = –1 in (i), we get –20 + 15 = A(–1 + 3) ⇒ –5 = 2A ⇒ A $= -\frac{5}{2}$Putting x = –3 in (i), we get –60 + 15 = B(–3 + 1) ⇒ –45 = –2B ⇒ B $= \frac{45}{2}$$\therefore I = \int\limits_{1}^{2} \left(5 + \frac{5}{2(x+1)} - \frac{45}{2(x+3)}\right) dx$$= \left[ 5x + \frac{5}{2} \log(x+1) - \frac{45}{2} \log(x+3) \right]_{1}^{2}$$= 5(2-1) + \frac{5}{2}[\log 3 - \log 2] - \frac{45}{2}[\log 5 - \log 4]$$= 5 - \frac{5}{2}\left(9\log\frac{5}{4} - \log\frac{3}{2}\right)$