Can someone please help me integrating this?
$$\displaystyle \int \frac{w\mathrm dw}{(5-3w)^{2/3}}$$
I tried substituting $5-3w = u$ and $-3\mathrm dw = \mathrm du$.
So $w = (u-5)/3$ and then we have
$$\int\frac{u-5}{3u^{2/3}}\mathrm du$$
How do I proceed?
$\endgroup$3 Answers
$\begingroup$Hint :
$$\int \frac{u-5}{3u^{2/3}} \,du =\frac{1}{3}\left( \int u^{1/3} \,du-5\int u^{-2/3} \,du\right)$$
$\endgroup$ 2 $\begingroup$Hint: $\dfrac{w}{(5-3w)^{2/3}} = \dfrac{5-(5-3w)}{3(5-3w)^{2/3}}=\dfrac{5}{3}\cdot (5-3w)^{-2/3}-\dfrac{1}{3}\cdot (5-3w)^{1/3}$
$\endgroup$ 1 $\begingroup$$$ \int \frac{w}{(5-3w)^{2/3}}dw \; $$
Let $u=5-3w$, then $w=\frac{5-u}{3}$ and $dw = \frac{-du}{3}$. Using these substitutions we get:
$$ \int \frac{\frac{5-u}{3}}{u^{2/3}}du \;\; = \frac{5}{3}\int \frac{du}{u^{2/3}}-\frac{1}{3}\int\frac{du}{u^{1/3}} $$
$\endgroup$
