How do we calculate the integral $\int_{1}^{e}\frac{\ln(x + 1)}{2x + 2}\mathrm{d}x$?

Question

I am trying to solve the following integral: $$\int_1^e \frac{\ln(x+1)}{2x+2} dx$$ so I make a substitution $$ t \rightarrow x + 1$$ and so $$\frac{1}{2}\int_2^{e+1} \frac{\ln(t)}{t} dt$$ and substitution again $$ u \rightarrow \ln(t)$$

so therefore $$\frac{1}{2}\int_{\ln(2)}^{\ln(e+1)} u du$$

and I believe that the final result should be $$\frac{\ln(e+1)^2 - \ln(2)^2}{4}$$

but in the exam I have taken, the only solution was

$$\frac{\ln(2)^2}{4}$$ unfortunately, my professor is away and I cannot contact him so I am wondering, if the professor made a mistake with the solution. Thanks

Answer 1

You are right indeed as you can check next:

\begin{align*} \int_{1}^{e}\frac{\ln(x + 1)}{2x + 2}\mathrm{d}x & = \frac{1}{2}\int_{1}^{e}\ln(x + 1)\mathrm{d}(\ln(x + 1))\\\\ & = \frac{1}{2}\int_{\ln(2)}^{\ln(e + 1)}u\mathrm{d}u\\\\ & = \frac{1}{4}\left(\ln^{2}(e + 1) - \ln^{2}(2)\right) \end{align*}

Hopefully this helps!