current attempt in progress use a calculator or computer to evaluate the integral. round your answer to one…

Explanation:

Step1: Recall integral formula

The antiderivative of $\frac{1}{\sqrt{1 + x^{2}}}$ is $\ln(x+\sqrt{1 + x^{2}})$.

Step2: Apply fundamental theorem of calculus

$\int_{1}^{4}\frac{1}{\sqrt{1 + x^{2}}}dx=\left[\ln(x+\sqrt{1 + x^{2}})\right]_{1}^{4}=\ln(4+\sqrt{1 + 16})-\ln(1+\sqrt{1 + 1})$.

Step3: Use calculator

$\ln(4+\sqrt{17})-\ln(1 + \sqrt{2})\approx1.4$.

Answer:

$1.4$