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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 ${\displaystyle\int}\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}\,\mathrm{d}x$

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$$I=\int\frac x{\sqrt{x^2-(a+b)x+ab}}dx$$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 $a,b$.

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    $\begingroup$ Is the function $\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}$ equal to $\frac x{\sqrt{x^2-(a+b)x+ab}}$ ? $\endgroup$ – Nosrati May 14 at 8:31
  • $\begingroup$ @Nosrati They are equal over the entire domain of the former $\endgroup$ – Shubham Johri May 14 at 9:20
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Hint: You can rationalize integrand by using substitution $t=\sqrt\frac{x-a}{x-b}.$

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