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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
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Answer

Let $\text{I}=\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}+\int\limits^2_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Consider $\text{f(x)}=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$\text{f}(-\text{x})=\int\limits^2_{-2}\frac{3(-\text{x}^3)}{(-\text{x}^2)+|-\text{x}|+1}\text{ dx}=\int\limits^2_{-2}\frac{-3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=-\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=-\text{f(x)}$
$\therefore\ \text{I}_1=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Now, Consider
$\text{g(x)}=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$\text{g}(-\text{x})=\int\limits^{2}_{-2}\frac{2|-\text{x}|+1}{(-\text{x}^2)+|-\text{x}|+1}\text{ dx}=\text{g(x)}=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}=\text{g}(\text{x)}$
$\therefore\ \text{I}_2=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=2\int\limits^{2}_{0}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
$=2\int\limits^{2}_{0}\frac{2\text{x}+1}{\text{x}^2+\text{x}+1}\text{ dx}$ $\begin{bmatrix}|\text{x}|\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}\end{bmatrix}$
$=2\times\Big[\log(\text{x}^2+\text{x}+1)\Big]^2_0$
$=2\times\big(\log7-\log1\big)$
$=2\times\big(7-0\big)$
$=2\log7$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$=0+2\log7$
$=2\log7$

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