#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 40 Maths Textbook Solution.

$\frac{2}{\sqrt{7}}\left [ \tan ^{-1}\frac{5}{\sqrt{7}}-\tan ^{-1}\frac{1}{\sqrt{7}} \right ]$

Hint: Use indefinite formula and then put limits to solve this integral

Given:

$\int_{0}^{1}\frac{1}{2x^{2}+x+1}dx$

Solution:

$\int_{0}^{1} \frac{1}{2 x^{2}+x+1} d x=\int_{0}^{1} \frac{1}{2\left(x^{2}+\frac{x}{2}+\frac{1}{2}\right)} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x^{2}+\frac{x}{2}+\frac{1}{2}\right)} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{x^{2}+2 \cdot x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}+\frac{1}{2}} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}+\frac{1}{2}} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\frac{1}{16}+\frac{1}{2}} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1-8}{16}\right)} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{-7}{16}\right)} d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}+\left(\frac{\sqrt{7}}{4}\right)^{2}} d x$

Put

$x+\frac{1}{4}=u\Rightarrow dx=du$

When $x=0$ then

$u=\frac{1}{4}$

when $x=1$ then

$u=1+\frac{1}{4}=\frac{4+1}{4}=\frac{5}{4}$

Then

\begin{aligned} &\frac{1}{2} \int_{0}^{1} \frac{1}{2 x^{2}+x+1} d x=\frac{1}{2} \int_{\frac{1}{4}}^{\frac{5}{4}} \frac{1}{u^{2}+\left(\frac{\sqrt{7}}{4}\right)^{2}} d u \\ &=\frac{1}{2}\left[\frac{\frac{1}{\sqrt{7}}}{4} \tan ^{-1} \frac{u}{\frac{\sqrt{7}}{4}}\right]_{\frac{1}{4}}^{\frac{5}{4}} \\ &=\frac{1}{2} \times \frac{4}{\sqrt{7}}\left[\tan ^{-1} \frac{4 u}{\sqrt{7}}\right]_{\frac{1}{4}}^{\frac{5}{4}} \end{aligned}

\begin{aligned} &=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \frac{4 \times \frac{5}{4}}{\sqrt{7}}-\tan ^{-1} \frac{4 \times \frac{1}{4}}{\sqrt{7}}\right] \\ &=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1} \frac{1}{\sqrt{7}}\right] \end{aligned}