# How to find ∫10ln3(1+x)lnxxdx{\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx

I suspect it might exist because there are similar integrals having closed forms:

Thanks!

Start with integration by parts (IBP) by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields

Applying IBP again to evaluate the red integral by setting $u=\ln^2(1-x)$ and $dv=\dfrac{\ln^2 x}{x}\ dx$ yields

For the simplicity, let

Introduce a generating function for the generalized harmonic numbers for $|x|<1$

and the following identity

Let us integrating the indefinite form of the blue integral.

Therefore

Using the similar approach as calculating the blue integral, then

Hence

Combining altogether, we have

Continuing my answer in: A sum containing harmonic numbers $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, we have

Dividing $(1)$ by $x$ and then integrating yields

Evaluating the red integral using the same technique as the previous one yields

Evaluating the purple integral yields

Evaluating the green integral using IBP by setting $u=\ln x$ and $dv=\dfrac{\operatorname{Li}_3(x)}{x}\ dx$ yields

Evaluating the orange integral using IBP by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields

Applying IBP again to evaluate the maroon integral by setting $u=\operatorname{Li}_2(1-x)$ and

we have

We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.

Dividing generating function of $\color{blue}{\mathbf{H}^{(k)}(x)}$ by $x$ and then integrating yields

Repeating the process above yields

where it is easy to show by using IBP that

and

Now, all unknown terms have been obtained. Putting altogether to $(2)$, we have

The next step is finding the constant of integration. Setting $x=1$ to $(3)$ yields

Thus

and setting $x=\frac12$ to $(4)$ yields

Finally, we obtain

$[1]\$ Harmonic number
$[2]\$ Polylogarithm