Questions: ∫ from 1 to √2 of (x^2 + √x) / x^2 dx

Solution Steps

To solve the integral \(\int_{1}^{\sqrt{2}} \frac{x^{2}+\sqrt{x}}{x^{2}} dx\), we can simplify the integrand first. By breaking the fraction into two separate terms, we get \(\int_{1}^{\sqrt{2}} \left(1 + \frac{\sqrt{x}}{x^{2}}\right) dx\). We can then integrate each term separately.

Given the integral: \[ \int_{1}^{\sqrt{2}} \frac{x^{2} + \sqrt{x}}{x^{2}} \, dx \] we can simplify the integrand: \[ \frac{x^{2} + \sqrt{x}}{x^{2}} = 1 + \frac{\sqrt{x}}{x^{2}} = 1 + x^{-\frac{3}{2}} \]

We now integrate each term separately: \[ \int_{1}^{\sqrt{2}} \left(1 + x^{-\frac{3}{2}}\right) \, dx \] This can be split into two integrals: \[ \int_{1}^{\sqrt{2}} 1 \, dx + \int_{1}^{\sqrt{2}} x^{-\frac{3}{2}} \, dx \]

1. For the first integral: \[ \int_{1}^{\sqrt{2}} 1 \, dx = \left. x \right|_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

2. For the second integral: \[ \int_{1}^{\sqrt{2}} x^{-\frac{3}{2}} \, dx = \left. \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} \right|_{1}^{\sqrt{2}} = \left. -2 x^{-\frac{1}{2}} \right|_{1}^{\sqrt{2}} = -2 \left( (\sqrt{2})^{-\frac{1}{2}} - 1^{-\frac{1}{2}} \right) \] \[ = -2 \left( 2^{-\frac{1}{4}} - 1 \right) \]

Combining the results of the two integrals: \[ \sqrt{2} - 1 - 2 \left( 2^{-\frac{1}{4}} - 1 \right) \] Simplifying further: \[ \sqrt{2} - 1 - 2 \cdot 2^{-\frac{1}{4}} + 2 = \sqrt{2} + 1 - 2^{\frac{3}{4}} \]

Final Answer