Solve the indefinite integral $\int \frac{1}{x^{1/2}+x^{1/3}} dx$

Question

Solve $\int \frac{1}{x^{1/2}+x^{1/3}} dx$

My Attempt $$ \int \frac{1}{x^{1/2}+x^{1/3}} dx=\int\frac{dx}{x^{1/2}(1+x^{-1/6})} $$ Put $t=x^{1/2}\implies dt=\frac{dx}{2.x^{1/2}}\implies\frac{dx}{x^{1/2}}=2dt$ $$ \int \frac{1}{x^{1/2}+x^{1/3}} dx=2\int\frac{dt}{1+\frac{1}{t^{1/3}}} $$ What is the easiest method to solve this indefinite integral and how do I choose a proper substitution in such problems ?

Answer 1

Hint. Make the change of variable $$ x=u^6,\qquad dx=6u^5du, $$ then you get $$ \int \frac{1}{x^{1/2}+x^{1/3}} dx=6\int \frac{u^3}{u+1} du $$ which is standard to evaluate.

Answer 2

$$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=6\left(\frac{(x^{\frac{1}{6}}+1)^3}{3}-\frac{3(x^{\frac{1}{6}}+1)^2}{2}+3(x^{\frac{1}{6}}+1)-\ln \left|x^{\frac{1}{6}}+1\right|\right)+k, \,k\in \mathbb Z$$

Proof:

If I take this substitution $t=x^{\frac{1}{6}}$ we have:

$$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=\int \frac{6t^3}{t+1}dt$$

With another substitution $v=t+1$ we have: $$\int \frac{6t^3}{t+1}dt=6\cdot \int \:\frac{(v-1)^3}{v}dv=6\left(\int \:v^2dv-\int \:3vdv+\int \:3dv-\int \frac{1}{v}dv\right)=$$

Hence

$$=6\left(\frac{v^3}{3}-\frac{3v^2}{2}+3v-\ln \left|v\right|\right)+k=*$$ and considering the two substitutions we have:

$$*=6\left(\frac{(x^{\frac{1}{6}}+1)^3}{3}-\frac{3(x^{\frac{1}{6}}+1)^2}{2}+3(x^{\frac{1}{6}}+1)-\ln \left|x^{\frac{1}{6}}+1\right|\right)+k, \,k\in \mathbb Z$$.