Branch line and integrals. Not finding the correct answer.

$$ \int_{0}^{\infty } \frac{ln x}{x^{3/4}(1+x)}dx$$

Integrating the real axis from 0 to infinity, so go around a big circle and close from - infinity to 0, around 0 close the path with a little circle.

$x = r$ along the real axis and Re>0. $I = \int_{0}^{\infty } \frac{ln x}{x^{3/4}(1+x)}dx$

Along the big circle, the integral vanishes.

along the real axis and Re>0, $ \int_{\infty}^{0} \frac{e^{i \pi /2 }ln r}{r^{3/4}(1+r)}dr = - \int_{0}^{\infty} \frac{e^{i \pi /2 }ln r}{r^{3/4}(1+r)}dr = -iI$

Along the little circle it will diverge ...

Along a circle around minus one, it diverges too. So what should i do? I can't see any other point .

If you're willing to use Fourier series instead of complex analysis, we can derive a more powerful result.

In this question, we use Fourier series to show that for $0 < \alpha < 1$ $$I(\alpha)=\int_0^{+\infty} \frac {1}{x^{\alpha}(1+x)}dx = \frac \pi {\sin(\pi \alpha)}$$ So taking the derivative w.r.t. $\alpha$ yields, for $0<\alpha < 1$, $$\boxed{\int_0^{+\infty} \frac {\ln(x)}{x^{\alpha}(1+x)}dx = \frac {\pi^2\cos(\pi \alpha)} {\sin^2(\pi \alpha)}}$$ Evaluating at $\alpha = \frac 3 4$ gives $$\boxed{\int_0^{+\infty} \frac {\ln(x)}{x^{\frac 3 4}(1+x)}dx = -\pi^2 \sqrt{2}}$$