# is this definite integral correct?

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.

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

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.

\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}