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#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 38 Maths Textbook Solution.

$\frac{1}{5}log6+\frac{3}{\sqrt{5}}\tan ^{-1}\left ( \sqrt{5} \right )$

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

Given:

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

Solution:

\begin{aligned} &\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x=\int_{0}^{1}\left(\frac{2 x}{5 x^{2}+1}+\frac{3}{5 x^{2}+1}\right) d x \\ &=\int_{0}^{1} \frac{2 x}{5 x^{2}+1} d x+\int_{0}^{1} \frac{3}{5 x^{2}+1} d x \\ &=\frac{1}{5} \int_{0}^{1} \frac{5 \times 2 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{1}{5 x^{2}+1} d x \\ &=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{1}{5 x^{2}+1} d x \end{aligned}                        ..........(1)

Now,

$\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x$

Put

$5x^{2}+1=u\Rightarrow 10xdx=du$

$\begin{array}{ll} \frac{1}{5} \int_{0}^{1} \frac{1}{u} d u=\frac{1}{5}[\log u]_{0}^{1} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\int \frac{1}{x} d x=\log x\right]} \\ =\frac{1}{5}\left[\log \left(5 x^{2}+1\right)\right]_{0}^{1} &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\left[u=5 x^{2}+1\right]} \end{array}$

\begin{aligned} &=\frac{1}{5}\left[\log \left(5 \cdot 1^{2}+1\right)-\log \left(5.0^{2}+1\right)\right] \\ &=\frac{1}{5}[\log (5+1)-\log (0+1)] \\ &=\frac{1}{5}[\log 6-\log 1] \end{aligned}

$\left [ \because log\: 1=0 \right ]$

$=\frac{1}{5}log\: 6$                                                        .........................(2)

And

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

\begin{aligned} &=\frac{1}{\sqrt{5}}\left[\tan ^{-1} \sqrt{5} \cdot 1-\tan ^{-1} \sqrt{5} .0\right] \\ &=\frac{1}{\sqrt{5}}\left[\tan ^{-1} \sqrt{5}-\tan ^{-1} 0\right] \\ &=\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5})-0 \\ &=\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5}) \end{aligned}                                    .....................(3)

Put the value of the integrals from (2) and (3) in (1) then

\begin{aligned} &\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x=\frac{1}{5} \log 6+3 \cdot \frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5}) \\ &=\frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5}) \end{aligned}