For x0, let f(x) = _1^x t1+t,dt.

Question

For $x>0$, let $f(x) = \displaystyle\int_1^x \dfrac{\log t}{1+t}\,dt$. Then $f(x)+f\left(\dfrac1x\right)$ is equal to:

Accepted Answer

$f\left(\dfrac1x\right) = \int_1^{1/x} \dfrac{\ln t}{1+t}\,dt$. Let $t=\dfrac1z$, $dt = -\dfrac{1}{z^2}dz$

$f\left(\dfrac1x\right) = \int_1^x \dfrac{\ln z}{z(z+1)}\,dz$

$f(x)+f\left(\dfrac1x\right) = \int_1^x \dfrac{\ln z}{z}\,dz = \left[\dfrac{(\ln z)^2}{2}\right]_1^x = \dfrac{(\ln x)^2}{2}$

Solutions · 1