I found the method of partial fractions very laborious to solve this definite integral :
$$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx$$
Is there a simpler way to do this ?
Answer
Perhaps this is simpler.
Make the substitution $\displaystyle x^{2/3} = t$. Giving us
$\displaystyle \frac{2 x^{1/3}}{3 x^{2/3}} dx = dt$, i.e $\displaystyle x^{1/3} dx = \frac{3}{2} t dt$
This gives us that the integral is
$$I = \frac{3}{2} \int_{0}^{\infty} \frac{t}{1 + t^3} \ \text{d}t$$
Now make the substitution $t = \frac{1}{z}$ to get
$$I = \frac{3}{2} \int_{0}^{\infty} \frac{1}{1 + t^3} \ \text{d}t$$
Add them up, cancel the $\displaystyle 1+t$, write the denominator ($\displaystyle t^2 - t + 1$) as $\displaystyle (t+a)^2 + b^2$ and get the answer.
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