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Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 45 Maths Textbook Solution.

Answers (1)

Answer:

$\frac{57-\sqrt{3}}{5}$

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

Given:

$\int_{1}^{4}\frac{x^{2}+x}{\sqrt{2x+1}}dx$

Sol:

$\int_{1}^{4}\frac{x^{2}+x}{\sqrt{2x+1}}dx$

Putting

$\Rightarrow 2x+1=t^{2}$

$\Rightarrow 2dx=2tdt$

$\Rightarrow dx=tdt$

and When $x=4$ then

$t=3$

and when $x=4$ then

$t=3$

then

$\int_{\sqrt{3}}^{3} \frac{\left(\frac{t^{2}-1}{2}\right)^{2}+\left(\frac{t^{2}}{2}-\frac{1}{2}\right)}{\sqrt{t^{2}}} \cdot t d t=\int_{\sqrt{3}}^{3} \frac{\frac{1}{4}\left(t^{2}-1\right)^{2}+\frac{1}{2}\left(t^{2}-1\right)}{t} \cdot t d t$

$=\int_{\sqrt{3}}^{3}\left(\frac{1}{4}\left(t^{4}+1-2 t^{2}\right)+\frac{t^{2}-1}{2}\right) d t$

$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}+1-2 t^{2}+2 t^{2}-2}{4}\right) d t$

$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}-1}{4}\right) d t$

$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}}{4}-\frac{1}{4}\right) d t$

$=\frac{1}{4} \int_{\sqrt{3}}^{3} t^{4} d t-\frac{1}{4} \int_{\sqrt{3}}^{3} t^{0} d t$

$=\frac{1}{4}\left[\frac{t^{4+1}}{4+1}\right]_{\sqrt{3}}^{3}-\frac{1}{4}\left[\frac{t^{0+1}}{0+1}\right]_{\sqrt{3}}^{3}$

$=\frac{1}{4}\left[\frac{t^{5}}{5}\right]_{\sqrt{3}}^{3}-\frac{1}{4}[t]_{\sqrt{3}}^{3}$

$=\frac{1}{20}\left[t^{5}\right]_{\sqrt{3}}^{3}-\frac{1}{4}[t]_{\sqrt{3}}^{3}$

$=\frac{\left[(3)^{5}-(\sqrt{3})^{5}\right]}{20}-\frac{1}{4}[3-\sqrt{3}]$

$=\frac{1}{4}\left[\frac{243-9 \sqrt{3}}{5}-3+\sqrt{3}\right]$

$=\frac{1}{4}\left[\frac{243-9 \sqrt{3}-3 \times 5+5 \sqrt{3}}{5}\right]$

$=\frac{1}{4}\left[\frac{243-9 \sqrt{3}-15+5 \sqrt{3}}{5}\right]$

$=\frac{1}{4}\left[\frac{228-4 \sqrt{3}}{5}\right]$

$=\frac{228}{20}-\frac{\sqrt{3}}{5}$

$=\frac{114}{10}-\frac{\sqrt{3}}{5}$

$=\frac{57}{5}-\frac{\sqrt{3}}{5}$

$=\frac{57-\sqrt{3}}{5}$

Posted by

infoexpert21

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