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}=

Explanation:

Step1: Complete the square for the denominator

For the quadratic expression $x^{2}+2x + 37$, we use the formula $(a + b)^2=a^{2}+2ab + b^{2}$. Here $a = x$ and $2ab=2x$, so $b = 1$. Then $x^{2}+2x+37=(x + 1)^{2}-1 + 37=(x + 1)^{2}+36$. So $\frac{1}{x^{2}+2x + 37}=\frac{1}{(x + 1)^{2}+36}$.

Step2: Use the integral formula

The integral $\int\frac{dx}{(x + 1)^{2}+36}$ is in the form of $\int\frac{du}{u^{2}+a^{2}}$ where $u=x + 1$, $du=dx$ and $a = 6$. The formula for $\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C$.
Substituting $u=x + 1$ and $a = 6$ into the formula, we get $\int\frac{dx}{(x + 1)^{2}+36}=\frac{1}{6}\arctan(\frac{x + 1}{6})+C$.

Answer:

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