_1^2 5 x^2 x^2+4 x+3 d x - Vidyadip

Question

$\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$

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Answer

Let $I=\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$
$
=\int_1^2\left[\frac{5\left(x^2+4 x+3\right)-5(4 x+3)}{x^2+4 x+3}\right] d x
$
$
=\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right] d x
$
$
\therefore I=\int_1^2 5 d x-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x
$
Let $\frac{20 x+15}{x^2+4 x+3}=\frac{20 x+15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}$
$
\therefore 20 x+15=A(x+3)+B(x+1)
$
Put $x+1=0$, i.e. $x=-1$, we get
$
\begin{aligned}
& 20(-1)+15=A(2)+B(0) \\
& \therefore-5=2 A \quad \therefore A=-\frac{5}{2}
\end{aligned}
$
Put $x+3=0$, i.e. $x=-3$, we get
$
20(-3)+15=A(0)+B(-2)
$
$
\therefore-45=-2 B \quad \therefore B=\frac{45}{2}
$
$
\therefore \frac{20 x+15}{x^2+4 x+3}=\frac{\left(-\frac{5}{2}\right)}{x+1}+\frac{\left(\frac{45}{2}\right)}{x+3}
$
$\therefore$ from (1),
$
I=\int_1^2 5 d x-\int_1^2\left[\frac{\left(-\frac{5}{2}\right)}{x+1}+\frac{\left(\frac{45}{2}\right)}{x+3}\right] d x
$
$\begin{aligned} & =5 \int_1^2 1 d x+\frac{5}{2} \int_1^2 \frac{1}{x+1} d x-\frac{45}{2} \int_1^2 \frac{1}{x+3} d x \\ & =5[x]_1^2+\frac{5}{2}[\log |x+1|]_1^2-\frac{45}{2}[\log |x+3|]_1^2 \\ & =5(2-1)+\frac{5}{2}(\log 3-\log 2)-\frac{45}{2}(\log 5-\log 4) \\ & =5+\frac{1}{2}(5 \log 3-5 \log 2-45 \log 5+90 \log 2) \\ & =5+\frac{1}{2}(5 \log 3+85 \log 2-45 \log 5) .\end{aligned}$

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