Question

evaluate the integral.
int\frac{dx}{sqrt{121 + x^{2}}}
int\frac{dx}{sqrt{121 + x^{2}}}=square
(type an exact answer, using radicals as needed.)

Explanation:

Step1: Recall integral formula

We know that $\int\frac{dx}{\sqrt{a^{2}+x^{2}}}=\ln|x + \sqrt{x^{2}+a^{2}}|+C$. Here $a^{2}=121$, so $a = 11$.

Step2: Apply the formula

Substituting $a = 11$ into the formula, we get $\int\frac{dx}{\sqrt{121+x^{2}}}=\ln|x+\sqrt{x^{2}+121}|+C$.

Answer:

$\ln|x+\sqrt{x^{2}+121}|+C$