Indefinite integration: $\int x^{x^2+1}(2\ln x+1)dx$

Question

Find the value of the integral: $$\int x^{x^2+1}(2\ln x+1)dx.$$ My attempt: I tried by using integration by parts, but not working since $x^{x^2+1}$ keeps coming again and again.

Then I tried putting $x^{x^2+1}=t$, this also is not helpful. Can someone help me in solving this question?

Answer 1

For the integral $$\int x^{x^2+1} \, (2\ln x+1) \, dx$$ let $u = x^{2} \, \ln x$ for which $du = (2x \, \ln x + x ) \, dx$ and \begin{align} I &= \int x^{x^{2} + 1} \, (2 \, \ln x + 1) \, dx \\ &= \int e^{x^{2} \, \ln x} \, (2 x \, \ln x + x) \, dx \\ &= \int e^{u} \, du \\ &= e^{u} + c_{0} \\ &= e^{x^{2} \, \ln x} + c_{0} = x^{x^{2}} + c_{0}. \end{align} Hence $$\int x^{x^2+1} \, (2\ln x+1) \, dx = x^{x^{2}} + c_{0}.$$

Answer 2

Observe that

$$x^{x^2+1}(2\log x+1)=x^{x^2}(2x\log x+x)=x^{x^2}\left(\frac{d\,\log x^{x^2}}{dx}\right)=\frac{d\,x^{x^2}}{dx}$$

Thus,

$$\int x^{x^2+1}(2\log x+1)dx=x^{x^2}+C$$

Answer 3

Let $x^{x^2}=t\implies (x^{x^2}(2x)\ln x+x^2x^{x^2-1})dx=dt$ $$x^{x^2+1}(2\ln x+1)dx=dt$$ Hence, we have $$\int x^{x^2+1}(2\ln x+1)dx=\int dt$$$$=t+C$$ $$=x^{x^2}+C$$