$\frac{1}{x^2} \int xe^x dx$ without using integration by parts

Question

On a test i just had, i needed to solve a differential equation which lead me to having to find the result of

$$ \frac{1}{x^2}\int xe^x dx $$

I then attempted to do this integral without integration by parts, (i did this because i forgot about integration by parts, yes i know you can scold me later if you wish) this is my work

$$ x e^x = \sum_{k=0}^{\infty}\frac{x^{k+1}}{k!} \therefore $$

$$ \frac{1}{x^2}\int xe^x dx = \frac{1}{x^2}\sum_{k=0}^{\infty}\frac{x^{k+2}}{(k+2) k!} $$

I know the answer is $e^x(\frac{1}{x}-\frac{1}{x^2})$ but how do i go about finding this result from the summation? Edit:

I tried to manipulate in this way

$$ \frac{1}{x^2}+\frac{1}{x^2}\sum_{k=0}^{\infty}\frac{x^k (k-1)}{k!} $$ But this seems even worse.

Answer 1

We have $$\int_0^t e^{\lambda x}dx = \dfrac{e^{\lambda t}-1}{\lambda}$$ Differentiating with respect to $\lambda$, we obtain $$\int_0^t xe^{\lambda x}dx = \dfrac{\lambda t e^{\lambda t} - e^{\lambda t }+1}{\lambda^2}$$ Set $\lambda = 1$, to obtain $$\int_0^t xe^{x}dx = t e^{t} - e^{t }+1$$

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EDIT To complete your approach, note that $$\sum_{k=0}^{\infty} \dfrac{x^k(k-1)}{k!} = \sum_{k=0}^{\infty} \dfrac{kx^k}{k!} - \sum_{k=0}^{\infty} \dfrac{x^k}{k!} = \sum_{k=0}^{\infty} \dfrac{x^{k+1}}{k!} - \sum_{k=0}^{\infty} \dfrac{x^k}{k!} = xe^x-e^x$$ which is same as the previous method, except for a constant addition and now finish it off.

Answer 2

Leave out $1/x^2$, to be reinserted later: \begin{align} \int xe^x\,dx &=\sum_{k=0}^{\infty}\frac{x^{k+2}}{(k+2) k!}\\ &=\sum_{k=0}^{\infty}\frac{(k+1)x^{k+2}}{(k+2)!}\\ &=\sum_{k=2}^{\infty}\frac{(k-1)x^k}{k!}\\ &=\sum_{k=2}^{\infty}\frac{kx^k}{k!}-\sum_{k=2}^{\infty}\frac{x^k}{k!}\\ &=\sum_{k=2}^{\infty}\frac{x^k}{(k-1)!}-\sum_{k=2}^{\infty}\frac{x^k}{k!}\\ &=x\sum_{k=1}^{\infty}\frac{x^k}{k!}-\sum_{k=2}^{\infty}\frac{x^k}{k!}\\ &=x(e^x-1)-(e^x-x-1)\\ &=xe^x-x-e^x+x+1\\ &=e^x(x-1)+1 \end{align} The $1$ is absorbed in the arbitrary constant, so the final result is $$ \int xe^x\,dx=e^x(x-1)+c $$ Differentiating we indeed have $e^x(x-1)+e^x=xe^x$