Chapter 8, Problem 67CRE

To determine

(a)

To prove that $I_n=\frac{x^{n-1}}{n-1}-I_{n-2}$

To determine

(b)

Use the previous solution in (a) to calculate $I_n$ for $0\le n\le 5$

To determine

(c)

To show that, in general, $I_{2n+1}=\frac{x^{2n}}{2n}-\frac{x^{2n-2}}{2n-2}+\mathrm{...}+{(-1)}^{n-1}\frac{x^2}{2}+{(-1)}^n\frac{\ln (x^2+1)}{2}+C$

$I_{2n}=\frac{x^{2n-1}}{2n-1}-\frac{x^{2n-3}}{2n-3}+\mathrm{...}+{(-1)}^{n-1}x+{(-1)}^n{\tan }^{-1}x+C$