Double u-substitution Problem I have to find $$\int\limits_{0}^{1}x(x+a)^{1/p}\text{ d}x$$ and I'm told to use two substitutions, $u=(x+a)^{1/p}$ and $u=x+a$. 
However, I can't find the derivative of $u$, $\text{d}u$, to replace once I've done the substitution - could someone offer a bit of insight into what I'm missing here? Thank you!
 A: Hint:
$$\int_{0}^{1}x(x+a)^{1/p}\,dx = \int_{0}^{1}(x+a)^{1+1/p}\,dx-a\int_{0}^{1}(x+a)^{1/p}\,dx$$
and:
$$\int_{0}^{1}(x+a)^{\beta}\,dx = \left.\frac{(x+a)^{\beta+1}}{\beta+1}\right|_{0}^{1}=\frac{(1+a)^{\beta+1}-a^{\beta+1}}{1+\beta}.$$
A: As the others have suggested, we need to use the anti-power rule:
$$\int{x^a}dx=\frac{1}{a+1}x^{a+1}$$
Also as the comment has suggested, we may use $x=(x+a)-a$ to substitute the first $x$:
$$\int_{0}^{1}x(x+a)^{\frac{1}{p}}dx = \int_{0}^{1}[(x+a)-a](x+a)^{\frac{1}{p}}dx$$ 
This works because $(x+a)-a = x+a-a = x$. Let us continue with the expansion:
$$\int_{0}^{1}[(x+a)(x+a)^{\frac{1}{p}} - a(x+a)^{\frac{1}{p}}]dx$$ 
$$=\int_{0}^{1}(x+a)^{\frac{1}{p}+1}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$
Now we use the anti-power rule to continue: ($\frac{1}{p}+1=\frac{1+p}{p}$)
$$\int_{0}^{1}(x+a)^{\frac{1+p}{p}}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$
$$=[(1+\frac{1+p}{p})(x+a)^{1+\frac{1+p}{p}} - a((1+\frac{1}{p})(x+a)^{1+\frac{1}{p}}]|_{0}^{1}$$
$$=\frac{1+2p}{p}(1+a)^{\frac{1+2p}{p}}-\frac{1+2p}{p}(a)^{\frac{1+2p}{p}}-a[\frac{p+1}{p}(1+a)^{\frac{p+1}{p}}-\frac{p+1}{p}(a)^{\frac{p+1}{p}}]$$
$$=\left(\frac{1}{p}+2\right)(1+a)^{\frac{1}{p}}(1+a)^2 - \left(\frac{1}{p}+1\right)(1+a)^{\frac{1}{p}}(1+a)a - \left(\frac{1}{p}+2\right)a^{\frac{1}{p}}a^2 + \left(\frac{1}{p}+1\right)a^{\frac{1}{p}}a^2$$
$$=(1+a)^{\frac{1}{p}}(1+a)^2 \left( \left(\frac{1}{p}+2\right) - \left(\frac{a}{1+a}\right) \right) - a^{\frac{1}{p}}a^2$$
$$=(1+a)^{\frac{1}{p}}(1+a)\left( \frac{1+a+2p+ap}{p} \right) - a^{\frac{1}{p}}a^2$$
