By using the substitution \(u=1/x\),
show that for
\(b>0\)
\[
\int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1)} \d x =0 \,.
\]
By using the substitution \(u=1/x\),
show that for \(b>0\),
\[
\int_{1/b}^b \frac{\arctan x}{x} \d x = \frac{\pi \ln b} 2\,.
\]
By using the result \( \displaystyle \int_0^\infty \frac 1 {a^2+x^2} \d x = \frac {\pi}{2 a} \) (where \(a > 0\)),and a substitution of the form \(u=k/x\), for suitable \(k\), show that
\[
\int_0^\infty \frac 1 {(a^2+x^2)^2} \d x = \frac {\pi}{4a^3 }
\, \ \ \ \ \ \ (a > 0).
\]
Solution:
\begin{align*}
&& I &= \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1} \d x \\
u = 1/x, \d u = -1/x^2 \d x: &&&= \int_{u=b}^{u=1/b} \frac{1/u \ln(1/u)}{(a^2+u^{-2})(a^2u^{-2}+1)} (- \frac{1}{u^2}) \d u \\
&&&= \int_{1/b}^b \frac{-u\ln u}{(a^2u^2+1)(a^2+u^2)} \d u \\
&&&= -I \\
\Rightarrow && I &= 0
\end{align*}
\(\,\) \begin{align*}
&& I &= \int_{1/b}^b \frac{\arctan x}{x} \d x \\
u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=b}^{u=1/b} \frac{\arctan \frac1u}{\frac1u} \frac{-1}{u^2} \d u \\
&&&= \int_{1/b}^b \frac{\arctan \frac1u}{u} \d u \\
\Rightarrow && 2I &= \int_{1/b}^b \frac{\arctan x + \arctan \frac1x}{x} \d x \\
&&&= \int_{1/b}^b \frac{\frac{\pi}2}{x} \d x \\
&&&= \pi \ln b \\
\Rightarrow && I &= \frac{\pi}{2} \ln b
\end{align*}