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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the definite integral in Exercise:
$\int\limits_{1}^{2}\frac{5\text{x}^{2}}{\text{x}^{2}+4\text{x}+3}\text{dx}$
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Answer

$\text{Let}\ \text{I}=\int\limits_{1}^{2}\frac{5\text{x}^{2}}{\text{x}^{2}+4\text{x}+3}\text{dx}$ $\text{Dividing}\ 5\text{x}^{2}\ \text{by}\ \text{x}^{2}+4\text{x}+3,\text{we}\ \text{obtain}$ $\text{I}=\int\limits_{1}^{2}\left\{5-\frac{20\text{x}+15}{\text{x}^{2}+4\text{x}+3}\right\}\text{dx}$ $=\int\limits_{1}^{2}5\text{dx}-\int^{2}\limits_{1}\frac{20\text{x}+15}{\text{x}^{2}+4\text{x}+3}\text{dx}$ $=[5\text{x}]^{2}_{1}-\int\limits_{1}^{2}\frac{20\text{x}+15}{\text{x}^{2}+4\text{x}+3}\text{dx}$ $\text{I}=5-\text{I},\ \text{where}\ \text{I}=\int\limits_{1}^{2}\frac{20\text{x}+15}{\text{x}^{2}+4\text{x}+3}\text{dx}$ $\text{consider}\ \text{I}_{1}\int\limits_{1}^{2}\frac{20\text{x}+15}{\text{x}^{2}+4\text{x}+8}\text{dx}$ $\text{Let}\ 20\text{x+}15=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^{2}+4\text{x}+3)+\text{B}$ $=2\text{Ax}+(4\text{A}+\text{B})$Equating the coefficiants of x and constant term, we obtain
$\text{A}=10\ \text{and}\ \text{B}=-25$ $\Rightarrow\text{I}_{1}=10\int\limits_{1}^{2}\frac{2\text{x}+4}{\text{x}^{2}+4\text{x}+3}\text{dx}-25\int\limits_{1}^{2}\frac{\text{dx}}{\text{x}^{2}+4\text{x}+3} $ $\text{Let}\ \text{x}^{2}+4\text{x}+3=\text{t}$ $\Rightarrow(2\text{x}+4)\text{dx}=\text{dt}$ $\Rightarrow\text{I}_{1}=10\int\frac{\text{dt}}{t}-25\int\frac{\text{dx}}{(\text{x}+2)^{2}-1^{2}}$ $=10\log\text{t}-25\bigg[\frac{1}{2}\log\bigg(\frac{\text{x}+2-1}{\text{x}+2+1}\bigg)\bigg]$ $=\big[10\log(\text{x}^{2}+4\text{x}+3)\big]^{2}_{1}-25\bigg[\frac{1}{2}\log\bigg(\frac{\text{x}+1}{\text{x}+3}\bigg)\bigg]_{1}^{2}$ $=\big[10\log15-10\log8\big]-25\bigg[\frac{1}{2}\log\frac{3}{5}-\frac{1}{2}\log\frac{2}{4}\bigg]$ $=\big[10\log(5\times3)-10\log(4\times2)\big]\frac{-25}{2}\big[\log3-\log5-\log2+\log4\big]$ $=\big[10\log5+10\log3-10\log4-10\log2]-\frac{25}{2}\big[\log3-\log5-\log2+\log4\big]$ $=\bigg[10+\frac{25}{2}\bigg]\log5+\bigg[-10-\frac{25}{2}\bigg]\log4+\bigg[10-\frac{25}{2}\bigg]\log3+\bigg[-10+\frac{25}{2}\bigg]\log2$ $=\frac{45}{2}\log5-\frac{45}{2}\log4-\frac{5}{2}\log3+\frac{5}{2}\log2$ $=\frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2}$Substituting the value of $\text{I}_1$ in (1), we obtain
$\text{I}=5-\bigg[\frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2}\bigg]$ $=5-\frac{5}{2}\bigg[9\log\frac{5}{4}-\log\frac{3}{2}\bigg]$

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