Calculate the following integral: $$\int \frac{x + \sqrt{x}}{\sqrt[4]{x} + \sqrt{x} + 1} dx$$
Here is what I did so far:
The expression in the integral can be rewritten as: $$\int xx^{-1/4} \frac{x^{-1/2} + x^{-1}}{1 + x^{1/4} +x^{-1/4}} dx$$
By substituting $x^{1/4}$ with a variable $t$, we get $dt = \frac{4}{5} xx^{-1/4}dx$. Now, lets apply this to our integral: $$\int \frac{5}{4} \frac{t^{-2} + t^{-4}}{1 + t + t^{-1}} dt = \frac{5}{4} \int \frac{t^2 + 1}{t^3(t^2 + t + 1)} dt$$
This all I did so far. I haven't succeded to solve the last integral yet.
Thank you!
After the change of variable you suggested I got : $$\class{steps-node}{\cssId{steps-node-1}{4}}{\displaystyle\int}\dfrac{u^5\left(u^2+1\right)}{u^2+u+1}\,\mathrm{d}u$$ Indeed $u=x^{1/4}$, so $du=\frac{dx}{4x^{3/4}}\Rightarrow dx=4x^{3/4}du=4u^3du$ and : $$\int \frac{x + \sqrt{x}}{\sqrt[4]{x} + \sqrt{x} + 1} dx=\int \frac{u^4 + u^2}{u + u^2 + 1} 4u^3du=\class{steps-node}{\cssId{steps-node-1}{4}}{\displaystyle\int}\dfrac{u^5\left(u^2+1\right)}{u^2+u+1}\,\mathrm{d}u$$
Then you should do long division and obtain : $$-{\displaystyle\int}\dfrac{1}{u^2+u+1}\,\mathrm{d}u+{\displaystyle\int}u^5\,\mathrm{d}u-{\displaystyle\int}u^4\,\mathrm{d}u+{\displaystyle\int}u^3\,\mathrm{d}u-{\displaystyle\int}u\,\mathrm{d}u+{\displaystyle\int}1\,\mathrm{d}u$$ You should be able to finish. Let me know if you still face difficulties.