Indefinite Integral

Examples

Proof that the following equation holds for all $n \in N$ via induction:

= I_{n} \int x^{n} (lo g x)^{n} d x = = S_{n} k = 0 \sum n \frac{( - 1 ) ^{k} \cdot n !}{( n - k )! \cdot ( n + 1 ) ^{k + 1}} \cdot x^{n + 1} \cdot (lo g x)^{n - k}

Start: Let $n = 0$ , then:

= I_{0} \int 1 d x = x = = S_{0} \frac{1}{1} \cdot x \cdot 1 = x

Assumption:

I_{n} = S_{n}

Step:

I_{n + 1} = \int x^{n + 1} (lo g x)^{n + 1} d x = \frac{x ^{n + 2}}{n + 2} (lo g x)^{n + 1} - \int \frac{n + 1}{n + 2} \frac{x ^{n + 2}}{x} (lo g x)^{n} d x = \frac{x ^{n + 2}}{n + 2} (lo g x)^{n + 1} - \frac{n + 1}{n + 2} \int x^{n + 1} (lo g x)^{n} d x = \frac{x ^{n + 2}}{n + 2} (lo g x)^{n + 1} - \frac{n + 1}{n + 2} [\frac{x ^{n + 2}}{n + 2} (lo g x)^{n} - \frac{n}{n + 2} \int x^{n + 1} (lo g x)^{n - 1} d x] = \frac{x ^{n + 2}}{n + 2} (lo g x)^{n + 1} - \frac{n + 1}{( n + 2 ) ^{2}} x^{n + 2} (lo g x)^{n} + \frac{( n + 1 ) n}{( n + 2 ) ^{2}} \int x^{n + 1} (lo g x)^{n - 1} d x = \frac{x ^{n + 2}}{n + 2} (lo g x)^{n + 1} - \frac{n + 1}{( n + 2 ) ^{2}} x^{n + 2} (lo g x)^{n} + \frac{( n + 1 ) n}{( n + 2 ) ^{3}} x^{n + 2} (lo g x)^{n - 1} - \frac{( n + 1 ) n ( n - 1 )}{( n + 2 ) ^{3}} \int x^{n + 1} (lo g x)^{n - 2} d x ⋮ = \frac{x ^{n + 2}}{n + 2} [(lo g x)^{n + 1} - \frac{n + 1}{n + 2} (lo g x)^{n} + \frac{( n + 1 ) n}{( n + 2 ) ^{2}} (lo g x)^{n - 1} - \dots + (- 1)^{n + 1} \frac{( n + 1 )!}{( n + 2 ) ^{n + 1}}] = = S_{n + 1} k = 0 \sum n + 1 \frac{( - 1 ) ^{k} \cdot ( n + 1 )!}{( n + 1 - k )! \cdot ( n + 2 ) ^{k + 1}} \cdot x^{n + 2} \cdot (lo g x)^{n + 1 - k} .

q.e.d.