Evaluate $$\displaystyle\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$$
I am tryed to integrate by parts with $du = 1$ and $v=\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)$
$$vu – \displaystyle\int v du =x\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)-{\displaystyle\int}\dfrac{x\left(\frac{1}{2\sqrt{x-b}}+\frac{1}{2\sqrt{x-a}}\right)}{\sqrt{x-b}+\sqrt{x-a}}\,\mathrm{d}x$$
Which further simplifies to
$$x\ln\left(\sqrt{x-b}+\sqrt{x-a}\right) – \frac12{\displaystyle\int}\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}\,\mathrm{d}x$$
I am stuck here.
Best Answer
$$I=\int\frac x{\sqrt{x^2-(a+b)x+ab}}dx$$You can write $x=0.5[2x-(a+b)+(a+b)]$,$$I=\int\frac{0.5[2x-(a+b)+(a+b)]}{\sqrt{x^2-(a+b)x+ab}}dx\\=0.5\left[\int\frac{2x-(a+b)}{\sqrt{x^2-(a+b)x+ab}}dx+\int\frac{(a+b)}{\sqrt{\left[x-\frac{(a+b)}2\right]^2-(a+b)^2/4+ab}}dx\right]$$The first integral reduces to $\int\frac{du}{\sqrt u}$ where $u=x^2-(a+b)x+ab$, while the second integral reduces to one of the standard forms $\int\frac{dv}{\sqrt{v^2-k^2}}$ or $\int\frac{dv}{\sqrt{v^2+k^2}}$ where $v=\left[x-\frac{(a+b)}2\right]$ depending on the value of $a,b$.