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 $v=\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)$


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

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 $a,b$.

  • 1
    $\begingroup$ 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 '19 at 8:31
  • $\begingroup$ @Nosrati They are equal over the entire domain of the former $\endgroup$ – Shubham Johri May 14 '19 at 9:20

Hint: You can rationalize integrand by using substitution $t=\sqrt\frac{x-a}{x-b}.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.