# Calculate a given integral

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!

• You need to do partial fractions. But the constants that Wolfram Alpha shows are not nice. The antiderivative it shows is not that bad, though. Commented Dec 10, 2016 at 18:33

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.