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This integral

$$\int xe^{6x} dx$$

is easily solved with integration by parts, but is it possible to solve with U substitution or some other method.

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  • $\begingroup$ It can certainly be changed into a different indefinite integral by the use of $u$-substitution, e.g. $u=e^{6x}$. I don't think this makes the problem easier though. $\endgroup$ – hardmath Dec 16 '19 at 14:53
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Let $xe^{6x} = t \implies e^{6x}(6x+1)dx =dt$

  • So first multiply and divide the given integrand by $6$.
  • Then add and subtract $1$

$$\int xe^{6x}dx = \frac16\int6xe^{6x}dx = \frac16\int(6x+1)e^{6x}dx -\frac16\int e^{6x}dx = \frac16t-\frac1{36}e^{6x}+c\\ = \frac16xe^{6x}-\frac1{36}e^{6x}+c\\=\frac{1}{36}(6x-1)e^{6x}+c$$

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    $\begingroup$ This is basicly per partes. $\endgroup$ – Aqua Dec 16 '19 at 14:56
  • $\begingroup$ @Aqua Yeah but it doesn't involve the formula $\int udv = uv -\int vdu$ directly. $\endgroup$ – Ak19 Dec 16 '19 at 15:03
  • $\begingroup$ @Aqua: I disagree, there is a true $t$-substitution, though you need a trick to solve it. $\endgroup$ – Yves Daoust Dec 16 '19 at 15:08
  • $\begingroup$ This can be restated as use of the Ansatz $\int xe^{6x}dx=(Ax+B)e^{6x}+C$, whence we differentiate to solve for $A,\,B$. If that's parts, any proof is parts. $\endgroup$ – J.G. Dec 16 '19 at 16:53
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This method is how to kill a fly with a gun, anyway it's working. You need to find a primitive of $xe^{6x}$. Search it as $p(x)e^{6x}$ where $p(x)$ is a polynomial of first degree. This is justified by observing that if you have a function like $$f(x)=p(x)e^{kx}$$ it is $$f'(x)=(p'(x)+kp(x))e^{kx}$$ and $q(x)=p'(x)+kp(x)$ has the same degree as $p(x)$. In our case it must be

$$(p(x)e^{6x})'=p'(x)e^{6x}+6p(x)e^{6x}=xe^{6x}$$ for each $x \in \mathbb R$. This means that $$p'(x)+6p(x)=x$$ for each $x \in \mathbb R$. Now, put $p(x)=ax+b$ and you get $$a+6ax+6b=x$$. Therefore $a=\frac{1}{6}$ and $b=-\frac{1}{36}$ so that an primitive is $$F(x)=(\frac{1}{6}x-\frac{1}{36})e^{6x}.$$ I tell once again: I did it just for sake of find an alternative way, and not because this is a good method. Integration by parts is very much better.

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Let

$$I(\alpha) = \int_{-\infty}^x e^{\alpha t}dt = \frac1\alpha e^{\alpha x}$$

and recognize,

$$\int xe^{\alpha x} dx = I'(\alpha) + C = \left(\frac x\alpha - \frac1{\alpha^2}\right)e^{\alpha x}+ C$$

Thus,

$$\int xe^{6 x} dx= \left(\frac x6 - \frac1{36}\right)e^{6x} + C$$

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Through power series:

Let's remember that:

$ e^x = \sum\limits_{n=0}^\infty \frac{x^n}{n!} $

Then,

$\int xe^{6x} dx = \int x \sum\limits_{n=0}^\infty \frac{(6x)^n}{n!} dx$

$\int xe^{6x} dx = \sum\limits_{n=0}^\infty \int \frac{6^n x^{n+1}}{n!} dx$ =

$ = \sum\limits_{n=0}^\infty \frac{6^n \int x^{n+1}}{n!} dx = $

$ \sum\limits_{n=0}^\infty \frac{6^n x^{n+2}}{n!(n+2)} dx = $

Therefore,

$ \int xe^{6x} dx = x^2 \sum\limits_{n=0}^\infty \frac{(6x)^n}{n!(n+2)} dx $

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Some general methods using power series can be used for such problems. Let $\, f(x) := \exp(6x)\,$ and $\, g(x) := \int x f(x)\,dx.\,$ Assuming an algebraic relation between $\,x,\, f(x),\,$ and $\,g(x),\,$ use an algebraic relation finding tool with series expansions up to $O(x^{12})$ to get $$ 0 = 1 - f(x) - 36\,g(x) + 6\,xf(x). $$ Solving this linear equation for $\,g(x)\,$ gives the solution $$ g(x) = (1-f(x))/36 + x f(x)/6$$ plus a constant, of course. This method works because the integral $\,g(x)\,$ has an algebraic relation with both $\,x\,$ and $\,f(x).\,$

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