Calculate the integral: integral from 1 to 4 (1+

Calculate the integral: $\int_1^4 \frac{1+\sqrt{x}}{x^2} \, dx$.

We will use the Newton-Leibniz formula after finding the antiderivative.

Step 1: Split the integrand

\[\int_1^4 \frac{1+\sqrt{x}}{x^2} \, dx = \int_1^4 \frac{1}{x^2} \, dx + \int_1^4 \frac{\sqrt{x}}{x^2} \, dx\]

Step 2: Rewrite using exponents

\[= \int_1^4 x^{-2} \, dx + \int_1^4 x^{1/2 - 2} \, dx = \int_1^4 x^{-2} \, dx + \int_1^4 x^{-3/2} \, dx\]

Step 3: Find the antiderivatives

Using the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$:

\[\int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x}\]

\[\int x^{-3/2} \, dx = \frac{x^{-1/2}}{-1/2} = -\frac{2}{\sqrt{x}}\]

Step 4: Combine the antiderivatives

\[F(x) = -\frac{1}{x} - \frac{2}{\sqrt{x}}\]

Step 5: Apply the Newton-Leibniz formula

\[\int_1^4 \frac{1+\sqrt{x}}{x^2} \, dx = \left[-\frac{1}{x} - \frac{2}{\sqrt{x}}\right]_1^4\]

Step 6: Evaluate at the upper limit $x = 4$

\[F(4) = -\frac{1}{4} - \frac{2}{2} = -\frac{1}{4} - 1 = -\frac{5}{4}\]

Step 7: Evaluate at the lower limit $x = 1$

\[F(1) = -1 - 2 = -3\]

Step 8: Calculate the definite integral

\[\int_1^4 \frac{1+\sqrt{x}}{x^2} \, dx = F(4) - F(1) = -\frac{5}{4} - (-3) = -\frac{5}{4} + \frac{12}{4} = \frac{7}{4}\]

Verification:

We can verify our antiderivative by differentiation:

\[\frac{d}{dx}\left[-\frac{1}{x} - \frac{2}{\sqrt{x}}\right] = \frac{1}{x^2} + x^{-3/2} = \frac{1}{x^2} + \frac{\sqrt{x}}{x^2} = \frac{1+\sqrt{x}}{x^2}\]

Geometric interpretation:

The integral $\int_1^4 \frac{1+\sqrt{x}}{x^2} \, dx$ represents the area under the curve $y = \frac{1+\sqrt{x}}{x^2}$ from $x = 1$ to $x = 4$. At $x=1$: $y=2$; at $x=4$: $y=\frac{3}{16}=0.1875$. The function is positive and decreasing on $[1,4]$, and as $x$ increases the term $\frac1{x^2}$ dominates the decay.

Final answer:

\[\boxed{\int_1^4 \frac{1+\sqrt{x}}{x^2} \, dx = \frac{7}{4}}\]

Or in decimal form: $\frac{7}{4} = 1.75$