Indefinite integral: $\int x(1+2\ln^2x)\,dx$

Question

I'm blocked, since I do not know how to start integrating the following integral:

$$\int x(2(\ln(x))^2+1)\,dx$$

It would be a good idea integrate by parts?

Answer 1

Let $\ln x=t\implies x=e^t,dx=e^t\,dt$

$$\int x(2(\ln x)^2+1)\ dx=\int e^{2t}2t^2\ dt+\int e^{2t} \ dt$$

Now for the first part use integrate by parts

or use reduction formula:

$$\frac{d(t^ne^{at})}{dt}=at^ne^{at}+nt^{n-1}e^{at}$$

$$a\int t^ne^{at} \, dt=t^ne^{at}-n\int t^{n-1}e^{at} \, dt$$

Answer 2

With substitution $\ln x=\dfrac12v$ we have $$\int x\left(2\ln^2x+1\right)\,dx=\dfrac14\int e^v(v^2+2)\,dv$$ now use integrating by parts!

Edit: Using the formula $$\color{blue}{\int pe^x\,dx=e^x(p-p'+p''-\cdots)}$$ where $p$ is a polynomial, so \begin{align} \int e^v(v^2+2)\,dv &= e^v\left((v^2+2)-(2v)+(2)\right) \\ &= e^v(v^2-2v+4) \\ &= e^{2\ln x}(4\ln^2x-4\ln x+4) \\ &= 4x^2\ln^2x-4x^2\ln x+4x^2 \end{align}

Answer 3

with Integration by parts we get $$x^2(\ln(x))^2-x^2\ln(x)+x^2+C$$