Int (3x^3.5)/(3+x^2.5) dx
I tried using integration by parts and the equation uv-int (vdu) and im not getting the right answer. Could someone please show me the steps on how to do this question?
Have you considered making the u-substitution $\displaystyle u=\sqrt{x}$? Then $\displaystyle dx=2u\,du$, and your integral becomes
$\displaystyle \int\frac{3x^{7/2}}{3+x^{5/2}}\,dx=\int\frac{6u^8}{3+u^5}\,du$
$\displaystyle =\int{6u^3}-\frac{18u^3}{3+u^5}\,du$
The first term is easily integrated, leaving the rational function $\displaystyle \frac{18u^3}{3+u^5}$ to be integrated.
Using the fifth roots of unity, one can factor the denominator above into real linear and quadratic factors:
$\displaystyle (u^5+3)=(u+3^{1/5})(u^2-3^{1/5}\frac{1-\sqrt5}{2}u+3^{2/5})(u^2-3^{1/5}\frac{1+\sqrt5}{2}u+3^{2/5})$
One can then use partial fractions to separate the above into rational functions with linear or quadratic denominators, which you should be able to integrate.
--Kevin C.