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

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

Answers (1)

Answer:

                \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}

 

 

 

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