Evaluate $\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}} d x$
$y=6 \int \frac{x+\frac{5}{6}}{\sqrt{6+x-2 x^{2}}} d x$
$y=\frac{6}{-4} \int \frac{-4\left(x+\frac{5}{6}\right)}{\sqrt{6+x-2 x^{2}}} d x$
$y=-\frac{3}{2} \int \frac{-4 x-\frac{10}{3}+1-1}{\sqrt{6+x-2 x^{2}}} d x$
$y=-\frac{3}{2} \int \frac{-4 x+1}{\sqrt{6+x-2 x^{2}}} d x-\frac{3}{2} \int \frac{-\frac{10}{3}-1}{\sqrt{6+x-2 x^{2}}} d x$
$A=-\frac{3}{2} \int \frac{-4 x+1}{\sqrt{6+x-2 x^{2}}} d x$
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 $y=A+B$
$y=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+C$