#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 75 Maths Textbook Solution.

$-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+c$

Hint:

To solve the given solution we will use the partial fraction.

Given:

$\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}}$

Solution:

$\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}}$

$=\int \frac{\frac{-8}{2}(-4 x+1)+\left(\frac{13}{2}\right)}{\sqrt{6+x-2 x^{2}}} d x$

$=-\frac{3}{2}\left[\int \frac{1-4 x}{\sqrt{6+x-2 x^{2}}} d x+\frac{13}{2} \int \frac{1}{\sqrt{6+x-2 x^{2}}} d x\right.$

$=-\frac{3}{2}\left[\int \frac{d t}{\sqrt{t}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3+\left(\frac{x}{2}\right)-x^{2}}} d x\right.$

$=-\frac{3}{2} \times 2 \sqrt{t}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3-\left(x^{2}-\left(\frac{x}{2}\right)+\frac{1}{16}-\frac{1}{16}\right)}}$

$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3+\left(\frac{1}{16}\right)-\left(x-\left(\frac{1}{4}\right)\right)^{2}}}$

$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{7}{4}\right)^{2}-\left(x-\left(\frac{1}{4}\right)\right)^{2}}} d x$

$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{x-\left(\frac{1}{4}\right)}{\frac{7}{4}}\right)+c$

$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+c$