Problem 38 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{1-x}{1-\sqrt{x}} d x$$

Step-by-Step Solution

Answer

Answer: The integral of \(\frac{1 - x}{1 - \sqrt{x}} dx\) is \(x + \frac{2x\sqrt{x}}{3} + 2C\), where \(C\) is the constant of integration.

Step 1: Let u be the square root of x

Let's substitute \(u\) for \(\sqrt{x}\). $$u = \sqrt{x}$$

Step 2: Find du

To find \(du\), differentiate the \(u\) with respect to \(x\) and multiply by \(d x\). $$\frac{d u}{d x} = \frac{1}{2 \sqrt{x}} \implies d u = \frac{1}{2 \sqrt{x}} d x$$ Now, note that this relates \(du\) to \(\sqrt{x}\) and \(dx\), allowing us to substitute into the original integral.

Step 3: Replace x and dx with u

From \(u = \sqrt{x}\), we know that \(x = u^2\). Also, we have \(d u = \frac{1}{2 \sqrt{x}} d x\), which implies \(d x = 2u du\). Substitute these expressions into the integral: $$\int \frac{1 - x}{1 - u} (2u du) = 2 \int \frac{1-u^2}{1-u} u du$$

Step 4: Simplify the integrand

We can simplify the integrand by dividing the numerator and denominator by \((1-u)\): $$2 \int \frac{(1 - u^2) / (1 - u)}{(1 - u) / (1-u)} u du = 2 \int (1+u)u du$$

Step 5: Integrate with respect to u

Now, integrate the simplified expression with respect to \(u\). $$2 \int (1+u)u du = 2\int (u+u^2) du$$ Now integrate each term separately: $$= 2\left[\frac{u^2}{2} + \frac{u^3}{3} + C\right]$$

Step 6: Substitute back x

Since \(u = \sqrt{x}\), we can substitute back to get the final answer. $$= 2\left[\frac{(\sqrt{x})^2}{2} + \frac{(\sqrt{x})^3}{3} + C\right]$$ Simplify the expression: $$= x + \frac{2x\sqrt{x}}{3} + 2C$$ So, the final answer is: $$\int \frac{1-x}{1-\sqrt{x}} d x = x + \frac{2x\sqrt{x}}{3} + 2C$$