# Evaluate the following integrals:

Question:

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

Solution:

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

$y=\frac{6}{-4} \int \frac{-4\left(x+\frac{5}{6}\right)}{\sqrt{6+x-2 x^{2}}} d x$

$y=-\frac{3}{2} \int \frac{-4 x-\frac{10}{3}+1-1}{\sqrt{6+x-2 x^{2}}} d x$

$y=-\frac{3}{2} \int \frac{-4 x+1}{\sqrt{6+x-2 x^{2}}} d x-\frac{3}{2} \int \frac{-\frac{10}{3}-1}{\sqrt{6+x-2 x^{2}}} d x$

$A=-\frac{3}{2} \int \frac{-4 x+1}{\sqrt{6+x-2 x^{2}}} d x$

Let, $6+x-2 x^{2}=t$

Differentiating both side with respect to t

$(1-4 x) \frac{d x}{d t}=1 \Rightarrow(1-4 x) d x=d t$

$A=-\frac{3}{2} \int \frac{1}{\sqrt{t}} d t$

$A=-\frac{3}{2} 2 \sqrt{t}+c_{1}$

Again, put $t=6+x-2 x^{2}$

$A=-3 \sqrt{6+x-2 x^{2}}+c_{1}$

$B=-\frac{3}{2} \int \frac{-\frac{10}{3}-1}{\sqrt{6+x-2 x^{2}}} d x$

$B=\frac{13}{2} \int \frac{1}{\sqrt{6+x-2 x^{2}}} d x$

Make perfect square of quadratic equation

$6+x-2 x^{2}=2\left(\left(\frac{7}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}\right)$

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

Use formula $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}$

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

$B=\frac{13}{2 \sqrt{2}} \sin ^{-1} \frac{4 x-1}{7}+c_{2}$

The final solution of the question is $y=A+B$

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