Calculate the following integral: $\int \sqrt[3]{1+x\ln{x}} \cdot (1+\ln{x}) dx$

Question

I have to calculate the following integral

$$ \int \sqrt[3]{1+x\ln{x}} * (1+\ln{x}) dx$$

I have thought about using the following notation:

$$ t = {1+x\ln(x)} \Rightarrow x\ln{x} = t-1 $$

But here I get stuck as I'm not completely sure how to get $x$ alone, my only solution would be

$$ x^x = e^{t-1} $$ Not sure how that would actually help me tho.

How do I get that x alone and also , am I on the right track with this notation?

Answer 1

$$t = {1+x\ln(x)} \iff dt=(\ln x+1) dx$$

$$\int (1+\ln{x}) \sqrt[3]{1+x\ln{x}} dx=\int \sqrt[3]{t} dt=\frac{3t^{4/3}}{4}+C=\frac{3}{4}(1+\ln x)^{4/3}+C$$

And here's the amazing result

$$\int (1+\ln{x}) \sqrt[3]{1+x\ln{x}} dx=\frac{3}{4} (1+\ln{x}) \sqrt[3]{1+x\ln{x}}+C$$

Answer 2

The substitution you're doing is good! Just observe that, from $t=1+x\log x$, we get $$ dt=(1+\log x)\,dx $$ and you're almost done.