Integrate $\int_1^e \frac{\cosh(ln(x))}{x}\, dx$ I have no idea on how to do this. I've tried many times using the substitution rule but I don't get anywhere. Please help.
 A: Hint:
$$ \cosh{\log{x}} = \frac{e^{\log{x}}+e^{-\log{x}}}{2} = \frac{1}{2}\left(x+\frac{1}{x}\right). $$
Alternative: $y=\log{x}$. Then $dy/dx = 1/x$.
A: The integral can be written as $$\int_1^e \! \cosh(\ln x) \cdot \frac1x\ \mathrm{d}x$$
Let $u=\ln x$. So $\dfrac1x \ \mathrm{d}x=\mathrm{d}u$. The new bounds become $\left[\ln 1 , \ln e\right]$ so the integral is 
\begin{align*}
\int_0^1 \! \cosh(u) \ \mathrm{d}u &= \bigg[\sinh u\bigg]^{1}_0\\
&= \sinh 1
\end{align*}
A: $$\int_1^e \frac{\cosh(\ln(x))}{x}\, \text{d}x=$$
$$\int_1^e \frac{x^2+1}{2x^2}\, \text{d}x=$$
$$\frac{1}{2}\int_1^e \frac{x^2+1}{x^2}\, \text{d}x=$$
$$\frac{1}{2}\int_1^e \left(\frac{1}{x^2}+1\right)\, \text{d}x=$$
$$\frac{1}{2}\left(\int_1^e \frac{1}{x^2}\, \text{d}x+\int_1^e 1\, \text{d}x\right)=$$
$$\frac{1}{2}\left(\left[-\frac{1}{x}\right]_{1}^{e}+\left[x\right]_{1}^{e}\right)=$$
$$\frac{1}{2}\left(\left(-\frac{1}{e}\right)-\left(-\frac{1}{1}\right)+e-1\right)=$$
$$\frac{1}{2}\left(-\frac{1}{e}+1+e-1\right)=$$
$$\frac{1}{2}\left(-\frac{1}{e}+e\right)=$$
$$\frac{e^2-1}{2e}$$
