evaluate $int\frac{dx}{cosh(2x + 3)coth(2x + 3)}$

Explanation:

Step1: Use substitution

Let $u = 2x+3$, then $du=2dx$ and $dx=\frac{1}{2}du$. The integral becomes $\frac{1}{2}\int\frac{du}{\cosh(u)\coth(u)}$.

Step2: Simplify the integrand

Recall that $\coth(u)=\frac{\cosh(u)}{\sinh(u)}$. So $\frac{1}{\cosh(u)\coth(u)}=\frac{\sinh(u)}{\cosh^{2}(u)}$. The integral is now $\frac{1}{2}\int\frac{\sinh(u)}{\cosh^{2}(u)}du$.

Step3: Use substitution again

Let $t = \cosh(u)$, then $dt=\sinh(u)du$. The integral becomes $\frac{1}{2}\int\frac{dt}{t^{2}}$.

Step4: Integrate

We know that $\int t^{- 2}dt=-\frac{1}{t}+C$. So $\frac{1}{2}\int\frac{dt}{t^{2}}=-\frac{1}{2t}+C$.

Step5: Back - substitute

Substitute back $t = \cosh(u)$ and $u = 2x + 3$. We get $-\frac{1}{2\cosh(2x + 3)}+C$.

Answer:

$-\frac{1}{2\cosh(2x + 3)}+C$