# Compute $\int_{1}^{e}\frac{1+\log x}{2x}dx$

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?

• If you want to change the variable back, you need to adjust the limits of integration as well. – ryagami Jan 21 '15 at 0:21
• Of course, there is no need to do that. You just need to calculate $\left.\frac{1}{2}t^2\right|_1^2$. – ryagami Jan 21 '15 at 0:22

$\int\limits_{1}^{2}t dt=\dfrac{t^2}{2}|_{1}^{2}=\dfrac{3}{2}$. You also have a factor of $\dfrac{1}{2}$ sitting outside, and so you get $\dfrac{3}{4}$
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}