Evaluate the Integral \(\int_{1}^{a^{2}} \frac{\ln x}{\sqrt{x}(x+a)} dx\)

This problem was posed at Mathematics Stack Exchange, here is my solution.

Let \(t=\sqrt{x}\)
I = \int\limits_{1}^{a^{2}} \frac{\ln x}{\sqrt{x}(x+a)} dx
= 4 \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt

Integrating by parts, we have
I_{1} &= \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt \\
&= \frac{\ln t}{\sqrt{a}} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) \Big|_{1}^{a}
\, – \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt \\
&= \frac{\ln a}{\sqrt{a}} \tan^{-1}(\sqrt{a})
\, – \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt

Let \(y=t/ \sqrt{a}\)
I_{2} &= \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt
= \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{1}{y} \tan^{-1}(y) dy \\
&= \frac{i}{2} \left[ \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1-iy)}{y} dy
\, – \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1+iy)}{y} dy \right] \\
&= \frac{i}{2} \left[ \mathrm{Li}_{2}(-iy) – \mathrm{Li}_{2}(iy) \right] \Big|_{1/\sqrt{a}}^{\sqrt{a}} \\
&= \frac{i}{2} \left( \left[ \mathrm{Li}_{2}(-i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{i}{\sqrt{a}}\right) \right]
– \left[ \mathrm{Li}_{2}(i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{-i}{\sqrt{a}}\right) \right] \right) \\
&= \frac{i}{2} \left( \left[ -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(i\sqrt{a}) \right]
– \left[ -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(-i\sqrt{a}) \right] \right) \\
&= \frac{\pi}{4} \ln a
a. \(\tan^{-1}(y) = \frac{i}{2} [\ln (1-iy) – \ln (1+iy)]\)

b. Dilogarithm function
\mathrm{Li}_{2}(z) = -\int_{0}^{z} \frac{\ln (1-x)}{x} dx

c. Use the identity
\mathrm{Li}_{2}(z) + \mathrm{Li}_{2}(1/z) = -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(-z)

d. \(\ln (\pm iz) = \ln z \pm i\pi /2\)

Now we have
I = 4I_{1} = \frac{4}{\sqrt{a}} (\ln a) \tan^{-1}(\sqrt{a}) \, – \frac{\pi}{\sqrt{a}} \ln a

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