# Find the integral of $\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$

Evaluate $$\displaystyle\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$$.

I am tryed to integrate it by parts by taking $$du = 1$$ and $$v=\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)$$

Therefore,

$$vu - \displaystyle\int v du =x\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)-{\displaystyle\int}\dfrac{x\left(\frac{1}{2\sqrt{x-b}}+\frac{1}{2\sqrt{x-a}}\right)}{\sqrt{x-b}+\sqrt{x-a}}\,\mathrm{d}x$$

Which further simplifies to $$=x\ln\left(\sqrt{x-b}+\sqrt{x-a}\right) - 0.5{\displaystyle\int}\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}\,\mathrm{d}x$$

I am stuck here. I need help solving $${\displaystyle\int}\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}\,\mathrm{d}x$$

$$I=\int\frac x{\sqrt{x^2-(a+b)x+ab}}dx$$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 $$a,b$$.
• Is the function $\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}$ equal to $\frac x{\sqrt{x^2-(a+b)x+ab}}$ ? – Nosrati May 14 '19 at 8:31
Hint: You can rationalize integrand by using substitution $$t=\sqrt\frac{x-a}{x-b}.$$