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$\int_{0}^{1} \frac{5 x^{2}}{x^{2}+4 x+3}$


Let $I=\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} d x$

Dividing $5 x^{2}$ by $x^{2}+4 x+3$, we obtain

$I=\int_{1}^{2}\left\{5-\frac{20 x+15}{x^{2}+4 x+3}\right\} d x$

$=\int_{1}^{2} 5 d x-\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3} d x$

$=[5 x]_{1}^{2}-\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3} d x$

$I=5-I_{1}$, where $I=\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3} d x$    ….(1)

Consider $I_{1}=\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+8} d x$

Let $20 x+15=\mathrm{A} \frac{d}{d x}\left(x^{2}+4 x+3\right)+\mathrm{B}$

$=2 \mathrm{~A} x+(4 \mathrm{~A}+\mathrm{B})$

Equating the coefficients of x and constant term, we obtain

$A=10$ and $B=-25$

$\Rightarrow I_{1}=10 \int_{1}^{2} \frac{2 x+4}{x^{2}+4 x+3} d x-25 \int_{1}^{2} \frac{d x}{x^{2}+4 x+3}$

Let $x^{2}+4 x+3=t$

$\Rightarrow(2 x+4) d x=d t$

$\Rightarrow I_{1}=10 \int \frac{d t}{t}-25 \int \frac{d x}{(x+2)^{2}-1^{2}}$

$=10 \log t-25\left[\frac{1}{2} \log \left(\frac{x+2-1}{x+2+1}\right)\right]$

$=\left[10 \log \left(x^{2}+4 x+3\right)\right]_{1}^{2}-25\left[\frac{1}{2} \log \left(\frac{x+1}{x+3}\right)\right]_{1}^{2}$

$=[10 \log 15-10 \log 8]-25\left[\frac{1}{2} \log \frac{3}{5}-\frac{1}{2} \log \frac{2}{4}\right]$

$=[10 \log (5 \times 3)-10 \log (4 \times 2)]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10 \log 5+10 \log 3-10 \log 4-10 \log 2]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=\left[10+\frac{25}{2}\right] \log 5+\left[-10-\frac{25}{2}\right] \log 4+\left[10-\frac{25}{2}\right] \log 3+\left[-10+\frac{25}{2}\right] \log 2$

$=\frac{45}{2} \log 5-\frac{45}{2} \log 4-\frac{5}{2} \log 3+\frac{5}{2} \log 2$

$=\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}$

Substituting the value of I1 in (1), we obtain

$I=5-\left[\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\right]$

$=5-\frac{5}{2}\left[9 \log \frac{5}{4}-\log \frac{3}{2}\right]$



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