Definite integral

Let $a>1$. Evaluate the integral:

$$\int_1^{a^2} \frac{\ln x}{\sqrt{x}(x+a)}\, {\rm d}x$$

Solution



 A first approach:

$$\begin{aligned}
\int_{1}^{a^2}\frac{\ln x}{\sqrt{x}\left ( x+a \right )}\, {\rm d}x &\overset{u=\sqrt{x}}{=\! =\! =\!} 4\int_{1}^{a}\frac{\ln u}{u^2 +a}\, {\rm d}u \\
 &=\left [ 4 \frac{\arctan \frac{u}{\sqrt{a}}}{\sqrt{a}} \ln u \right ]_1^a - \frac{4}{\sqrt{a}}\int_{1}^{a}\frac{\arctan \frac{u}{\sqrt{a}}}{u}\, {\rm d}u \\
 &= \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1}^{a}\frac{\arctan \frac{u}{\sqrt{a}}}{u}\, {\rm d}u\\
 &\overset{y=u/\sqrt{a}}{=\! =\! =\! =\! =\!} \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\arctan y}{y}\, {\rm d}y \\
 & \overset{y=1/t}{=\! =\! =\!}  \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\arctan \frac{1}{t}}{t}\, {\rm d}t \\
 & =  \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\left [ \frac{\pi/2 - \arctan t }{t} \right ]\, {\rm d}t \\
 &= \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{\pi}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{{\rm d}t}{t} \\
 &= \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{\pi \ln a}{\sqrt{a}} \\
 & = -\ln a \left [ \frac{\pi - 4\arctan \sqrt{a}}{\sqrt{a}} \right ]
\end{aligned}$$

A second approach:

$$\begin{aligned}
\int_{1}^{a^2}\frac{\ln x}{\sqrt{x}(x+a)}\, {\rm d}x &\overset{u=\sqrt{x}} {=\! =\! =\!=\!}\;4\int_{1}^{a}\frac{\ln u}{u^2+a}\, {\rm d}u \\
 &\overset{u=\sqrt{a}t}{=\! =\! =\! =\!} \;4\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\ln \sqrt{a}t}{a(t^2+1)}\cdot \sqrt{a}\, {\rm d}t \\
 &= \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln \sqrt{a}+\ln t}{t^2+1}\, {\rm d}t\\
 &=\frac{2\ln a}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{{\rm d}t}{t^2+1}+ \cancelto{0}{\frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\ln t}{t^2+1}\, {\rm d}t}\\
 &=-\ln a \left [ \frac{\pi - 4\arctan \sqrt{a}}{\sqrt{a}} \right ]
\end{aligned}$$

A third approach:

$$\begin{aligned}
 \int_{1}^{a^2}\frac{\ln x}{\sqrt{x}\left ( x+a \right )}\, {\rm d}x&\overset{u=\sqrt{x}}{=\! =\! =\! =\!} \int_{1}^{a}\frac{2u \ln u^2}{u(u^2+a)} \, {\rm d}u \\
 &=4\int_{1}^{a}\frac{\ln u}{u^2+a}\, {\rm d}u  \\
 &\overset{u=a/t}{=}2 \ln a \int_{1}^{a}\frac{{\rm d}u}{u^2 +a}  \\
 &=2\ln a \left [ \frac{\arctan \frac{u}{\sqrt{a}}}{\sqrt{a}} \right ]_1^a \\
 &= 2\ln a \left [ \frac{\arctan \sqrt{a}-\arctan \frac{1}{\sqrt{a}}}{\sqrt{a}} \right ] \\
 & -\ln a \left [ \frac{\pi -4 \arctan \sqrt{a}}{\sqrt{a}} \right ]
\end{aligned}$$

In all of the above solutions we made use of the (basic) trigonometric equation:

$$\arctan x +\arctan \frac{1}{x}= \frac{\pi}{2}, \;\; x>0$$