#### Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 31 textbook solution.

Answer:-  $2\log_{e}7$

Hints:-  You must know the rules of integration .

Given:-  $\int_{-2}^{2} \frac{3 x^{3}+2|x|+1}{x^{2}+|x|+1} d x$

Solution :  $I=\int_{-2}^{2} \frac{3 x^{3}+2|x|+1}{x^{2}+|x|+1} d x$

\begin{aligned} &=\int_{-2}^{2} \frac{3 x^{3}}{x^{2}+|x|+1}+\int_{-2}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x\\ &\text { Odd function } \quad \text { Even function } \end{aligned}

\begin{aligned} &=2 \int_{0}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x=2 \log _{e} 7 \\ &\frac{\pi}{2} \\ &\int_{0}^{1}\left(\tan ^{-1}(x)+\tan ^{-1}(1-x)-\tan ^{-1}(1-x)-\tan ^{-1}(x)\right) d x \end{aligned}

$I=\frac{\pi}{2}$

\begin{aligned} &I=\frac{16}{15} \sqrt{2} \\ &{\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]} \\ &I=2 \sqrt{2}\left(\frac{4}{3}-\frac{4}{5}\right) \end{aligned}

\begin{aligned} &I=2 \int_{0}^{2} \frac{d t}{t} \\ &=2 \int_{0}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x \end{aligned}

Put $x^{2}+|x|+1=t$

$\begin{gathered} 2 x+1 d x=d t \\ I=2 \int_{0}^{2} \frac{d t}{t} \\ =2\left[\log _{e}|t|\right]_{0}^{2} \end{gathered}$

\begin{aligned} &=2\left[\log _{e}\left|x^{2}+\right| x|+1|\right]_{0}^{2} \\ &=2\left[\log _{e}(4+2+1)-\log (0+0+1)\right] \\ &=2\left[\log _{e} 7-\log _{e} 1\right] \\ &=2 \log _{e} 7 \end{aligned}