# Can $\int xe^{6x} dx$ be solved without integration-by-parts

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.

• 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. – hardmath Dec 16 '19 at 14:53

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 $$1$$

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

• This is basicly per partes. – Aqua Dec 16 '19 at 14:56
• @Aqua Yeah but it doesn't involve the formula $\int udv = uv -\int vdu$ directly. – Ak19 Dec 16 '19 at 15:03
• @Aqua: I disagree, there is a true $t$-substitution, though you need a trick to solve it. – Yves Daoust Dec 16 '19 at 15:08
• 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. – J.G. Dec 16 '19 at 16:53

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.

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,

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

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,

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

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 $$\,f(x).\,$$