Question

evaluate the following integral.
int\frac{dx}{x^{2}+2x + 37}

Explanation:

Step1: Complete the square in the denominator

We have $x^{2}+2x + 37=(x + 1)^{2}+36$. So the integral becomes $\int\frac{dx}{(x + 1)^{2}+36}$.

Step2: Use substitution

Let $u=x + 1$, then $du=dx$. The integral is $\int\frac{du}{u^{2}+36}$.

Step3: Recall the integral formula

The formula for $\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C$. Here $a = 6$. So $\int\frac{du}{u^{2}+36}=\frac{1}{6}\arctan(\frac{u}{6})+C$.

Step4: Substitute back

Substitute $u=x + 1$ back into the result. We get $\frac{1}{6}\arctan(\frac{x + 1}{6})+C$.

Answer:

$\frac{1}{6}\arctan(\frac{x + 1}{6})+C$