TITLE: Compute $\int_{1}^{e}\frac{1+\log x}{2x}dx$
QUESTION [0 upvotes]: I've been trying to solve this integral: $$\int_{1}^{e}\frac{1+\log x}{2x}dx$$
I used a new variable to solve this; 
$1+\log x = t$ therefore   $dx = x dt$, then I inserted this into the original equation and changed the $e$ and $1$:
$$\begin{align}\int_{1}^{2}\frac{t\ x}{2x}dt &= \int_{1}^{2}\frac{t}{2}dt = \frac12\int_{1}^{2}tdt \\&=
\frac12t^2\Bigg|_1^2 = \frac12(1+\log x)^2\Bigg|_1^2 \\&=\frac12(1+2\log x+\log^2x)|_1^2=(\frac12+\log x+\frac12\log^2x)|_1^2 \approxeq 0.933\end{align}$$
When you insert $1$ and $2$ into the equation, it does not equal $\frac34$ which is the result of this integral. 
Where did I go wrong?

REPLY [1 votes]: You may write
$$
\begin{align}
\int_{1}^{e}\frac{1+\log x}{2x}dx&=\int_{1}^{e}\frac{1}{2x}dx+\int_{1}^{e}\frac{\log x}{2x}dx\\\\
&=\left[\frac{1}{2} \log x\right]_1^e+\left[\frac{1}{4} \left(\log x\right)^2\right]_1^e\\\\
&=\frac{1}{2} +\frac{1}{4}\\\\
&=\frac{3}{4}.
\end{align}
$$
{"set_name": "stack_exchange", "score": 0, "question_id": 1112823}