Problem 30 - Essential Calculus

Question

Verify the integration formula. $$ \int(\ln u)^{n} d u=u(\ln u)^{n}-n \int(\ln u)^{n-1} d u $$

Step-by-Step Solution

Answer

The given integration formula \( \int (\ln u)^n du = u(\ln u)^n - n \int (\ln u)^{n-1} du \) has been verified successfully using integration by parts and a recursive approach.

Step 1: Define the parts in terms of integration by parts

Let \( u = (\ln x)^n \) and \( dv = dx \). Then we differentiate \( u \) and integrate \( dv \) to get \( du = n (\ln x)^{n-1} * 1/x dx \) and \( v = x \) respectively. According to the integration by parts formula, the right side of the given formula becomes \( u \int v dx - \int u'(\int v dx) dx \). Substituting \( u, du, \) and \( v \), we get \( (\ln x)^n * x - \int n (\ln x)^{n-1} * 1/x * x dx \). After simplifying, the equation becomes \( x(\ln x)^n - n \int (\ln x)^{n-1} dx \).

Step 2: Compare the results

As per our working in step 1, we found out that \( \int (\ln x)^n dx = x(\ln x)^n - n \int (\ln x)^{n-1} dx \). Looking at our original formula, \( \int (\ln u)^n du = u(\ln u)^n - n \int (\ln u)^{n-1} du \), they both match perfectly, confirming the given integration formula is correct.