Compute $$\int_0^\infty \frac{x^{1/3}}{(x+8)(x+1)^2}dx$$
I think I will have to consider the curve to be the boundary of a three-quarter of annulus of outer radius $R$ and inner radius $1/R$. (i.e. 1/R to R in the real axis, R to $-i$R via three-quarter circle, -iR to -i/R via imaginary axis, -i/R to 1/R via three-quarter circle)
However, I could not get a bound for inner radius and also could not work out the integral on the imaginary axis.
Thanks for the help.
Another approach.
Your integral: $$I=\int _0^{\infty }\frac{x^{\frac{1}{3}}}{\left(x+8\right)\left(x+1\right)^2}\:dx$$ $$=\frac{1}{49}\int _0^{\infty }\frac{x^{\frac{1}{3}}}{x+8}\:dx-\frac{1}{49}\int _0^{\infty }\frac{x^{\frac{1}{3}}}{x+1}\:dx+\frac{1}{7}\int _0^{\infty }\frac{x^{\frac{1}{3}}}{\left(x+1\right)^2}\:dx$$ We can now use the following $2$ identities: $$\int _0^{\infty }\frac{x^a}{x^b+c}\:dx=c^{\frac{a+1}{b}-1}\:\frac{\pi }{b}\csc \left(\pi \frac{a+1}{b}\right)$$and $$\int _0^{\infty }\frac{x^a}{\left(x^b+c\right)^2}\:dx=-\left(\frac{a+1}{b}-1\right)c^{\frac{a+1}{b}-2}\:\frac{\pi }{b}\csc \left(\pi \frac{a+1}{b}\right)$$ So, $$I=\frac{1}{49}\left(8^{^{\frac{1}{3}}}\pi \csc \left(\frac{4\pi }{3}\right)\right)-\frac{1}{49}\left(\pi \csc \left(\frac{4\pi }{3}\right)\right)+\frac{1}{7}\left(-\frac{1}{3}\pi \csc \left(\frac{4\pi }{3}\right)\right)$$ $$=-\frac{4\pi }{49\sqrt{3}}+\frac{2\pi }{49\sqrt{3}}+\frac{2\pi }{21\sqrt{3}}=\frac{8\pi }{147\sqrt{3}}$$ Thus: $$\boxed{I=\int _0^{\infty }\frac{x^{\frac{1}{3}}}{\left(x+8\right)\left(x+1\right)^2}\:dx=\frac{8\pi }{147\sqrt{3}}}$$