Question

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

Explanation:

Step1: Complete the square for the denominator

We know that \(x^{2}+2x + 37=(x^{2}+2x + 1)+36=(x + 1)^{2}+6^{2}\). So the integral becomes \(\int\frac{dx}{(x + 1)^{2}+6^{2}}\).

Step2: Use substitution

Let \(u=x + 1\), then \(du=dx\). The integral is \(\int\frac{du}{u^{2}+6^{2}}\).

Step3: Recall the integral formula

The integral formula \(\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C\) (where \(a = 6\) in our case). So \(\int\frac{du}{u^{2}+6^{2}}=\frac{1}{6}\arctan(\frac{u}{6})+C\).

Step4: Substitute back

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

Answer:

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