evaluate the integral.\n int\frac{dx}{sqrt{x^{2}-169}}=square \n(type an exact answer.)
evaluate the integral.\n int\frac{dx}{sqrt{x^{2}-169}}=square \n(type an exact answer.)
Answer
Explanation:
Step1: Recall integral formula
We use the formula $\int\frac{dx}{\sqrt{x^{2}-a^{2}}}=\ln|x + \sqrt{x^{2}-a^{2}}|+C$, where $a^{2}=169$, so $a = 13$.
Step2: Apply the formula
Substituting $a = 13$ into the formula, we get $\int\frac{dx}{\sqrt{x^{2}-169}}=\ln|x+\sqrt{x^{2}-169}|+C$.
Answer:
$\ln|x+\sqrt{x^{2}-169}|+C$