# How to integrate $(x^{1/2} + 5 x^{1/3})^{-1}$?

$$\int_0^1 \frac{dx}{x^\frac12+5x^\frac13}$$ Let $$u = 5+x^\frac16$$ I have change x into u, and find $$\int_5^6 \frac{7}{6u(u-5)^3}$$ Then, i don't know how to integrate it.

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Use partial fractions for the last integral –  Amr Dec 7 '12 at 11:35
I get something very different from ${7\over6u(u-5)^3}$ when I make the substitution. Can you show your work? –  Gerry Myerson Dec 7 '12 at 11:49

Use $x=t^6$ as a new substitution to integrate the first integral. So you have $$\int\frac{6t^5 dt}{t^2(t+5)}=\int\frac{6t^3dt}{t+5}=\int6\left( t^2-5t+25-\frac{125}{t+5}\right)dt$$ which is elementary. Whenever you have an integral like $$\int R(x,x^{\frac{p_1}{q_1}},...,x^{\frac{p_k}{q_k}})dx$$ you can use $x=t^{\text{lcm}({q_1},...{q_k})}$ as a substitution.

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I really love the bottom part of your answer. –  000 Dec 7 '12 at 12:05

$$\int \frac{1}{u(u-5)^3}du=\int\frac{A}{u}+\frac{B}{(u-5)}+\frac{C}{(u-5)^2}+\frac{D}{(u-5)^3}du.$$

Courtesy of partial fraction decomposition. :-)

Edit: It would appear your $u$ sub does not work. (Thanks, Gerry.) Nonetheless, I hope this is a helpful illustration of partial fraction decomposition.

Here is why your $u$ sub does not work: If $u=5+x^{\frac{1}{6}}$ and we wish to write $\dfrac{7}{6u(u-5)^3}$, we find the following: $$\frac{7}{6u(u-5)^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{6}})^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{2}})}=\frac{7}{6}\frac{1}{(5x^{\frac{1}{2}}+x^{\frac{2}{3}})}\ne \frac{1}{x^{\frac{1}{2}}+5x^{\frac{1}{3}}}.$$

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Putting $x=(t-5)^6, dx=6(t-5)^5$ and $x=0\implies t=5,x=1\implies t=6$

We get $$\int_0^1 \frac{dx}{x^\frac12+5x^\frac13}=\int_5^6 \frac{6(t-5)^3dt}{t} =6\int_5^6 (t^2-15t+75-125\frac1t)dt=6(\frac{t^3}3-15\frac{t^2}2+75t-125\log t)\mid_5^6$$ $$=2(6^3-5^3)-45(6^2-5^2)+450(6-5)-750\log\frac65$$

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This is the same substitution as OP uses (but carried out correctly). –  Gerry Myerson Dec 7 '12 at 12:01
@GerryMyerson Pardon my ignorance, but what do you mean by 'this is the same substitution'? –  000 Dec 7 '12 at 12:03
I mean your $x=(t-5)^6$ is the same as OP's $t=5+x^{1/6}$. Isn't it? –  Gerry Myerson Dec 7 '12 at 12:42
Sorry, comment above was meant for @Limitless. –  Gerry Myerson Dec 8 '12 at 6:34
@GerryMyerson Correct; I wasn't entirely sure you meant it like that. –  000 Dec 8 '12 at 6:45