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The answer of course: $$\int(2x^2+1)e^{x^2}\,dx=xe^{x^2}+C$$

But what kind of techniques we should use with problem like this ?

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@Babak Sorouh thanks . – Frank Aug 20 '12 at 0:23
up vote 6 down vote accepted

You can expand the integrand, and get


$$x\cdot 2x e^{x^2}+1\cdot e^{x^2}=$$

Note that $x'=1$ and that $(e^{x^2})'=2xe^{x^2}$ so you get

$$=x\cdot (e^{x^2})'+(x)'\cdot e^{x^2}=(xe^{x^2})'$$

Thus you integral is $xe^{x^2}+C$. Of course, the above is integration by parts in disguise, but it is good to develop some observational skills with problems of this kind.

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As @Peter note above, you can integrate this by separating the integrand and integrating by parts: \begin{align} \int(2x^2+1)e^{x^2}dx &=\int2x^2e^{x^2}dx+\int e^{x^2}dx\\ &=\int x(2xe^{x^2})dx+\int e^{x^2}dx\\ &= \int x\left(\frac{d}{dx}e^{x^2}\right)dx\ + \int e^{x^2}dx\\ &= xe^{x^2}-\int e^{x^2}dx+ \int e^{x^2}dx= xe^{x^2} + C \end{align}

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$\quad$:+)$\quad$ – amWhy Mar 13 '13 at 0:58

Look at the integral $$\int 2x^2e^{x^2}\, dx.$$ Try integrating by parts as follows. $u = x$, $dv = 2xe^{x^2}dx$, $v = e^{x^2}$, $du = dx$.

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My solution is essentially based on a theorem proved by Liouville in 1835, applied to your problem: There has to be a polynomial function of linear nature that satisfies an antiderivative of the form (ax+b) times the e-power from your integral. Left and right differentiation and "removing" the e-power gives (2x^2+1)=(2ax^2+2bx+a) Matching coefficients according to corresponding polynomial terms yields a = 1 and b = 0 I do admit, it may not be the most beautiful solution, but Liouville's theorem is the reason why the above methods actually come out good!

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