Question

evaluate the following integral.
int\frac{dx}{x^{2}+2x + 37}
rewrite the integrand by completing the square in the denominator.
\frac{1}{x^{2}+2x + 37}=\frac{1}{(x + 1)^{2}+36}
find the indefinite integral.
int\frac{dx}{x^{2}+2x + 37}=square

Explanation:

Step1: Recognize integral form

We know that $\int\frac{dx}{a^{2}+u^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C$. After completing the square, our integral is $\int\frac{dx}{(x + 1)^{2}+36}$, where $u=x + 1$ and $a = 6$.

Step2: Apply integral formula

Using the formula $\int\frac{dx}{a^{2}+u^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C$, we substitute $u=x + 1$ and $a = 6$ into it. So $\int\frac{dx}{(x + 1)^{2}+36}=\frac{1}{6}\arctan(\frac{x + 1}{6})+C$.

Answer:

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