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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 52 Maths Textbook Solution.

Answers (1)

Answer:

$I=5 \sqrt{x^{2}-9 x+20}+\frac{59}{2} \log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20}\right|+c$

Given:

$\int \frac{5 x+7}{\sqrt{(x-5)(x-4)}} d x$

Hint:

To solve this equation we have to do differentiation method.

Solution:

$I=\int \frac{5 x+7}{\sqrt{(x-5)(x-4)}} d x$

$I=\int \frac{5 x+7}{\sqrt{x^{2}-9 x+20}} d x$

Take 5/2 common from numerator and add and subtract 9,

$I=\frac{5}{2} \int \frac{2 x+\frac{14}{5}-9+9}{\sqrt{x^{2}-9 x+20}} d x$

$I=\frac{5}{2} \int \frac{(2 x-9)+\frac{14}{5}+9}{\sqrt{x^{2}-9 x+20}} d x$

$I=\frac{5}{2} \int \frac{(2 x-9)}{\sqrt{x^{2}-9 x+20}} d x+\frac{5}{2} \int \frac{\frac{59}{5}}{\sqrt{x^{2}-9 x+20}} d x$

$\text { put } \sqrt{x^{2}-9 x+20}=t$

$\frac{d t}{d x}=\frac{1}{2 \sqrt{x^{2}-9 x+20}} \times(2 x-9)$

$d x=\frac{d t \times 2}{2 x-9} \sqrt{x^{2}-9 x+20}$

$I_{1}=\frac{5}{2} \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} \times d t \frac{\left(2 \sqrt{x^{2}-9 x+20}\right)}{2 x-9}$

$I_{1}=5 \int d t$

$I_{1}=5 t$

$I_{1}=5 \sqrt{x^{2}-9 x+20}$

$I_{2}=\frac{59}{2} \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$

$=\frac{59}{2} \int \frac{1}{\sqrt{x^{2}-9 x+\frac{81}{4}-\frac{81}{4}+20}} d x$

$=\frac{59}{2} \int \frac{1}{\sqrt{x^{2}-9 x+\frac{81}{4}+\left(20-\frac{81}{4}\right)}} d x$

$=\frac{59}{2} \int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x$

$=\left[\because \int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right.$

$\text { let } x-\frac{9}{2}=u$

$d x=d u$

$I=\frac{59}{2} \int \frac{d u}{\sqrt{u^{2}-\left(\frac{1}{2}\right)^{2}}}$

$=\frac{59}{2} \log \left|u+\sqrt{u^{2}-\left(\frac{1}{2}\right)^{2}}\right|+c$

$=\frac{59}{2} \log \left|u+\sqrt{u^{2}-\left(\frac{1}{2}\right)^{2}}\right|+c$

$\text { Adding } I_{1} \text { and } I_{2}$

$I=5 \sqrt{x^{2}-9 x+20}+\frac{59}{2} \log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20}\right|+c$

Posted by

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