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