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!

  • 2
    $\begingroup$ Use the power rule and the chain rule. $\endgroup$ – TorsionSquid Sep 6 '15 at 19:22
  • 6
    $\begingroup$ Easier, I think, to write $x=(x+a)-a$ and split the integral into two. $\endgroup$ – lulu Sep 6 '15 at 19:23
  • $\begingroup$ For the substitution $u=x+a$, you can use $du=dx$ (since $\frac{du}{dx}=1$). For the substitution $u=(x+a)^{1/p}$, you have $x=u^p-a$ so $dx=pu^{p-1}du$. $\endgroup$ – user84413 Sep 6 '15 at 22:32

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: $$\int_{0}^{1}(x+a)^{\beta}\,dx = \left.\frac{(x+a)^{\beta+1}}{\beta+1}\right|_{0}^{1}=\frac{(1+a)^{\beta+1}-a^{\beta+1}}{1+\beta}.$$

  • $\begingroup$ Thank you, that actually helped get me started! $\endgroup$ – Faith Sep 10 '15 at 0:15

As the others have suggested, we need to use the anti-power rule:


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: ($\frac{1}{p}+1=\frac{1+p}{p}$)

$$\int_{0}^{1}(x+a)^{\frac{1+p}{p}}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$ $$=[(1+\frac{1+p}{p})(x+a)^{1+\frac{1+p}{p}} - a((1+\frac{1}{p})(x+a)^{1+\frac{1}{p}}]|_{0}^{1}$$ $$=\frac{1+2p}{p}(1+a)^{\frac{1+2p}{p}}-\frac{1+2p}{p}(a)^{\frac{1+2p}{p}}-a[\frac{p+1}{p}(1+a)^{\frac{p+1}{p}}-\frac{p+1}{p}(a)^{\frac{p+1}{p}}]$$ $$=\left(\frac{1}{p}+2\right)(1+a)^{\frac{1}{p}}(1+a)^2 - \left(\frac{1}{p}+1\right)(1+a)^{\frac{1}{p}}(1+a)a - \left(\frac{1}{p}+2\right)a^{\frac{1}{p}}a^2 + \left(\frac{1}{p}+1\right)a^{\frac{1}{p}}a^2$$ $$=(1+a)^{\frac{1}{p}}(1+a)^2 \left( \left(\frac{1}{p}+2\right) - \left(\frac{a}{1+a}\right) \right) - a^{\frac{1}{p}}a^2$$ $$=(1+a)^{\frac{1}{p}}(1+a)\left( \frac{1+a+2p+ap}{p} \right) - a^{\frac{1}{p}}a^2$$

  • $\begingroup$ That's awesome! Thank you for providing such a detailed solution, it was very helpful. $\endgroup$ – Faith Sep 10 '15 at 0:19

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