Problem 47 - Calculus Early Transcendentals

Question

Use integration by parts to derive the following reduction formulas. $$\int \ln ^{n} x d x=x \ln ^{n} x-n \int \ln ^{n-1} x d x$$

Step-by-Step Solution

Answer

Question: Derive the reduction formula for the function $$\int \ln^n x dx$$, where n is a positive integer. Answer: The reduction formula for the given function is $$\int \ln^n x dx = x \ln^n x - n \int \ln^{n-1} x dx$$.

Step 1: Identify the functions u and dv

The given function is: $$\int \ln^n x dx$$ Let \(u = \ln^n x\) and \(dv = dx\).

Step 2: Differentiate u to obtain du and integrate dv to obtain v

To differentiate u, we use the chain rule, where \(y = \ln x\). $$\frac{d}{dx} \ln^n x = n \ln^{n-1} x \frac{d}{dx} (\ln x) = \frac{n}{x} \ln^{n-1} x$$ So, \(du = \frac{n}{x} \ln^{n-1} x dx\). As for dv, we can integrate it directly to obtain v: $$v = \int dx = x$$

Step 3: Apply the integration by parts formula

Substitute u, du, and v into the integration by parts formula: $$\int \ln^n x dx = x \ln^n x - \int x \frac{n}{x} \ln^{n-1} x dx$$

Step 4: Combine the terms to derive the reduction formula

Simplify the equation obtained from Step 3: $$\int \ln^n x dx = x \ln^n x - n \int \ln^{n-1} x dx$$ Thus, we have derived the reduction formula: $$\int \ln^n x dx = x \ln^n x - n \int \ln^{n-1} x dx$$