# Problem integrating $\int\frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}dx$

I am not able to start the following integration , as in that the powers of $x$ are in fractional form . So it is very difficult for me to do substitution. $$\int\frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}dx$$

Can anybody please give me a start .

• – lab bhattacharjee Aug 21 '16 at 17:56
• Please use more distinctive titles in the future. I changed it now. – quid Aug 21 '16 at 19:25

$$=\int \frac{1}{x^{1/3}+x^{1/4}}dx$$

Factor out a $x^{1/4}$:

$$=\int \frac{1}{x^{1/4}(x^{1/12}+1)}dx$$

Let $u=x^{1/12}$. Then $du=\frac{1}{12} x^{-11/12} \,dx$ and $12x^{11/12}\,du=12u^{11}\,du=dx$. Also $x^{1/4}=u^3$. So we have:

$$=12 \int \frac{u^{11}}{u^3(u+1)}\,du$$

$$=12 \int \frac{u^8}{u+1}\, du$$

This is standard with another substitution $y=u+1$. The binomial theorem might be of great help.

• The thing to observe is that $12$ is the common denominator for $1/3$ and $1/4$. – GEdgar Aug 21 '16 at 16:58
• Here is a more elegant alternative: Subtract and Add $1$ to the numerator and use difference of powers to cancel the factor of $(x+1)$ from $x^8 -1$. – Jack Lam Aug 23 '16 at 10:35

Hint: take $u=\sqrt[12]{x}$. The integral becomes (why?) $$\int\frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}dx=12\int\frac{u^{8}}{u+1}du$$ and now doing long division we get $$12\int\frac{u^{8}}{u+1}du=12\int\left(u^{7}-u^{6}+u^{5}-u^{4}+u^{3}-u^{2}+u-1+\frac{1}{u+1}\right)du.$$

• Sorry , i tried but i am not getting from there – Koolman Aug 21 '16 at 16:19
• @koolman Now do long division. – Marco Cantarini Aug 21 '16 at 16:22
• @koolman as a thumb rule u can remember that whenever u see fraction power of x in deno.(of the integrand) L.C.M of the deno. of the fractions may be a good substitution. – Sujan Dutta Aug 21 '16 at 17:05
• @koolman : $$u = x^{1/12}$$ $$x^{1/4} = x^{3/12} = (x^{1/12})^3 = u^3$$ Similarly, you get $x^{1/3} = u^4$. Since $u = x^{1/12}$, you have: $$u^{12} = x, \text{ and so } 12u^{11}\,du = dx.$$ This makes the integral into $$\int \frac 1 {u^4 + u^3} (12u^{11}\,du).$$ Then simplify. $\qquad$ – Michael Hardy Aug 21 '16 at 17:40