Integrate indefinite integral

Question

How to integrate

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

It seems too long to integrate by parts (already tried without success) so i don’t what to do. Substitution doesn’t seem helpful neither. Thanks for your help.

Answer 1

Following Jaideep's first hint, let $t=x^2$. Then $$\int x\,(1+x^2)\,e^{x^2}\log x\,dx = \frac{1}{4}\int (1+t)\,e^t\log t\, dt.$$ (One factor of 2 is from $d(e^{x^2}) = 2x\,e^{x^2}$, the other is from $\log x^2 = 2\log x$.)

We try integrating by parts: \begin{align*} u &= (1+t)\log t & dv &= e^t\,dt \\ du &= \left(\log t + 1 + \frac{1}{t}\right) dt & v &= e^t \end{align*} \begin{align} \int (1+t)\,e^t\log t\, dt &= (1+t)\,e^t\log t - \int \left(\log t + 1 + \frac{1}{t}\right)e^t\,dt \\ &= (1+t)\,e^t\log t - e^t - \int e^t\log t\,dt-\int\frac{e^t}{t}\,dt. \end{align} Now it happens that the second integral there doesn't have an elementary expression. But if we integrate $\int e^t\log t\,dt$ by parts, we get a happy accident: \begin{align*} u &= \log t & dv &= e^t\,dt \\ du &= \frac{dt}{t} & v &= e^t \end{align*} $$ \int e^t\log t\,dt = e^t\log t - \int\frac{e^t}{t}\,dt. $$ We still can't evaluate that last integral, but it disappears when we substitute back: \begin{align*} \int (1+t)\,e^t\log t\,dt &= (1+t)\,e^t\log t - e^t - e^t\log t + C\\ &= e^t(t\log t - 1) + C \end{align*} and \begin{align*} \int x(1+x^2)e^{x^2}\log x\,dx &= \frac{e^{x^2}}{4}(x^2\log x^2 - 1) + C \\ &= \boxed{\frac{x^2\,e^{x^2}\log x}{2} - \frac{e^{x^2}}{4} + C}. \end{align*}

The broader lesson here, for math (and life in general?): if a problem breaks down into parts and some of them seem intractable, work on the other parts. Sometimes the nasty pieces take care of themselves!

Answer 2

Hint : Let $x^2=t$.

And then use that $$\int e^t(f(t)+f'(t)) {\rm d} t= e^t f(t) +\rm C$$