1
$\begingroup$

Show that the substitution $u=e^x$ converts $\int \frac{2+\ln(u)}{u^2} du$ into $\int \frac{2+x}{e^x} dx$Hence evaluate $\int_1^e \frac{2+\ln(u)}{u^2} du$I have tried every substitution available and still cannot come to an answer. I can find the answer via integration by parts but not through a substitution which I think is required. Thanks 

I don't think I'm supposed to use parts

$\endgroup$

closed as unclear what you're asking by Did, user7530, user230715, quid, 6005 Aug 30 '15 at 18:37

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is your problem? The substitution $u=e^x$ is given, so $\ln(u)=x$ and $du = u dx$. $\endgroup$ – gammatester Oct 22 '13 at 9:52
4
$\begingroup$

Hint: $$\int \frac{2+\ln(u)}{u^2}{du}$$ Now, Substitute $ u = e^x, \space du = e^xdx$ $$I = \int \frac{2+\ln e^x}{e^{2x}} e^x \, dx$$ $$I = \int \frac{2+x}{e^{2x}}e^x\,dx = \int \frac 2{e^x} \,dx + \int \frac x{e^x} \,dx$$ $$I = I_1 + I_2$$ Now you solve $I_1$ by simply integrating it and apply integration by parts in $I_2$. After that substitute back $ x = \ln(u)$ in the result.

$\endgroup$
0
$\begingroup$

Write your integrand as (2+x) Exp[-x]. Develop. Then, you have to integrate Exp[-x] + x Exp[-x]. The first one is simple. For the second, use integration by parts.

$\endgroup$
0
$\begingroup$

$$u = e^x\\ du = e^x dx\\ \int \frac{2+\ln(u)}{u^2} du= \int \frac{2+x}{e^{2x}} e^xdx= \int \frac{2+x}{e^x} dx\\ $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.