Question

Find the integral of $\sqrt{x^2-a^2}$ w.r.t $x$ and hence evaluate $\int \sqrt{x^2-8x+7}dx$

Answer 1

$I=\int \sqrt{x^2-a^2}dx=\sqrt{x^2-a^2}.x-\frac{1}{2}{(x^2-a^2)}^{1/2-1}\left(2x\right)$
$=x\sqrt{x^2-a^2}-\int \sqrt{x^2-a^2}-a^2\int \frac{1}{\sqrt{x^2-a^2}}$
$I=x\sqrt{x^2-a^2}-1-a^2\log |x+\sqrt{x^2-a^2}|$
$I=\int \sqrt{x^2-8x+7}dx=\int \sqrt{{(x-4)}^2-3^2}dx=\frac{x-4}{2}\sqrt{{(x-4)}^2-3^2}-\frac{9}{2}\log \left|(x-4)\sqrt{{(x-4)}^2-3^2}\right|+c$