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I have this definite integral $$ \int^{2}_{1} \frac{e^{1/x}}{x^4}dx$$ this is my attempt:

  • I used u-substitution. $u = 1/x$, and then $-du = 1/x^2 dx$
  • I rewrote $1/x^4$ as $(1/x^2) * (1/x^2)$
  • now, I rewrote the integral as $$ -\int^{1/2}_{1} u^2 e^u du$$
  • I used integration by parts technique twice, as follows: $g(x) = u^2, \space g'(x) = 2u\space du, \space and \space f(x) = e^u, \space f'(x) = e^u\space du$

$ e^u u^2 - 2\int^{1/2}_{1} e^u u\space du$

the second integration by parts is as follows: $g(x) = u$, $g'(x) = du$, and $f(x) = e^u$, $f'(x) = e^u\space du$

$ e^u u^2 - 2e^u u - \int^{1/2}_{1} e^u\space du$

$ -e^u u^2 + 2e^u u + e^u$

  • the result is the following: $-e^u u^2 + 2e^u u+ e^u$ (evaluated from $u =1$ to $u = 1/2$).

does it make sense? is it correct?

EDIT: perhaps, there's a problem with some signs, because before the integral I've put a minus sign, and because of that I should've changed signs in the result. fixed.

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    $\begingroup$ What's $t$ and how is it related to $x$ or $u$? $\endgroup$
    – Andrei
    Jan 26 at 18:10
  • $\begingroup$ sorry, it's a typo. I didn't notice that $\endgroup$ Jan 26 at 18:12
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    $\begingroup$ And can you put in the intermediate steps for the integration by parts? I think there are some factors of $2$ missing $\endgroup$
    – Andrei
    Jan 26 at 18:13
  • $\begingroup$ done, buty I think there are still a problem with signs $\endgroup$ Jan 26 at 18:22
  • $\begingroup$ You realize that you can troubleshoot your own work by differentiating your answer for the antiderivative. It helps to fact out the exponential. $\endgroup$ Jan 26 at 18:26

2 Answers 2

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I think you made an algebra mistake when integrating by parts. Starting with

$$\int u^2 e^u \; \mathrm{d}u,$$

your choice of $g = u^2, \mathrm{d}f = e^u\; \mathrm{d}u$ is good. This yeilds

$$\int u^2 e^u \; \mathrm{d}u = u^2 e^u - \int 2 u e^u \; \mathrm{d}u = u^2 e^u - {\color{blue} 2} \int u e^u \; \mathrm{d}u.$$

Next, we focus on $\int u e^u \; \mathrm{d}u$. Like you suggested, we pick $g = u, \mathrm{d} f = e^u \; \mathrm{d}u$. This gives us

$$\int u e^u \; \mathrm{d}u = u e^u - \int e^u \; \mathrm{d}u = ue^u - e^u + c.$$

Putting everything together, we have

$$\int u^2 e^u \; \mathrm{d}u = u^2 e^u - {\color{blue} 2}(ue^u - e^u + c) = e^u(u^2 - 2u + 2) + c.$$

I think you forgot to correctly distribute the factor of ${\color{blue} 2}$, and a minus sign.

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$$\begin{align}\int_1^2\frac{e^{1/x}}{x^4}dx&\xrightarrow{1/x\to u}-\int_{1}^{1/2}u^2e^udu\\&=\int_{1/2}^1u^2e^udu\\&=u^2e^u\Big|_{1/2}^1-2\int_{1/2}^1ue^udu\\&=u^2e^u\Big|_{1/2}^1-2ue^u\Big|_{1/2}^1+2\int_{1/2}^1e^udu\\&=(u^2-2u+2)e^u\Big|_{1/2}^1\end{align}$$

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