Problem 18 - Essential Calculus

Question

Find the indefinite integral by \(u\) -substitution. (Hint: Let \(u\) be the denominator of the integrand.) $$ \int \frac{\sqrt[3]{x}}{\sqrt[3]{x}-1} d x $$

Step-by-Step Solution

Answer

The solution to the integral is \( \frac{3}{2} (\sqrt[3]{x}-1)^2 x + x(\sqrt[3]{x}-1) + (\sqrt[3]{x})^3 -1 +C \).

Step 1: Set up the \(u\)-substitution

The denominator of the integrand is given as \(\sqrt[3]{x}-1\). According to the hint, we should let \(u\) be this value. Therefore, let \(u=\sqrt[3]{x}-1\). This implies that \(x=(u+1)^3\). We differentiate both sides with respect to \(x\) to find the substitution for \(dx\). On differentiating, we get \(dx = 3(u+1)^2 du\). We now substitute \(x\) and \(dx\) in the integral.

Step 2: Substitute \(x\) and \(dx\) in the integral

Substituting the values for \(x\) and \(dx\) in the integral gives: \( \int \frac{(u+1)}{u} \cdot 3(u+1)^2 du\).

Step 3: Simplify the integral

We factor out a 3 from the integral. Then split the fraction and rewrite the integral as two separate integrals. This gives: \(3 \int u(u+1)^2 du + 3 \int (u+1)^2 du\).

Step 4: Evaluate the integrals

Integrating each term separately using the power rule for integration yields the solution. The power rule of integration states that \(\int u^n du = \frac{1}{n+1}u^{n+1}\). On integrating each term, you will get \( \frac{3}{2} u^2 (u+1)^2 + u(u+1)^2 + (u+1)^3 + C \).

Step 5: Substitute back the value of \(u\)

Substitute back \(u = \sqrt[3]{x} - 1\) to get the answer in terms of \(x\). This gives: \( \frac{3}{2} (\sqrt[3]{x}-1)^2 x + x(\sqrt[3]{x}-1) + (\sqrt[3]{x})^3 -1 +C \).