( x-1)1+( x)^2,dx is equal to: *(reading the OCR'd numerator/denominator as ( x-1)^2[1+( x…

Question

$\displaystyle\int \dfrac{(\log x-1)}{1+(\log x)^2}\,dx$ is equal to:

*(reading the OCR'd numerator/denominator as $\int \dfrac{(\log x-1)^2}{[1+(\log x)^2]^2}\,dx$ per the worked solution)*

Accepted Answer

$\int \dfrac{(\log x-1)^2}{(1+(\log x)^2)^2}\,dx = \int \dfrac{1+(\log x)^2-2\log x}{[1+(\log x)^2]^2}\,dx = \int \left[\dfrac{1}{1+(\log x)^2}-\dfrac{2\log x}{[1+(\log x)^2]^2}\right]dx$

Put $\log x=t$, $x=e^t$: $I = \int e^t\left[\dfrac{1}{1+t^2}-\dfrac{2t}{(1+t^2)^2}\right]dt$

This is of the form $\int e^x(f(x)+f'(x))\,dx = f(x)e^x+c$ with $f(t)=\dfrac{1}{1+t^2}$:

$I = \dfrac{e^t}{1+t^2}+c = \dfrac{x}{1+(\log x)^2}+c$

Solutions · 1