Question

SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Answer 1

Required series is:

`\int{(-1)/(1+x^(2)) \, dx =-x+(x^(3))/(3)-(x^(5))/(5)+(x^(7))/(7)+.....`

Explanation:

Given that

`f'(x) = -(1)/(1 + x^(2))` ---(1)

We know that:

`(d)/(dx)(tan^(-1)x)=(1)/(1+x^(2))` ---(2)

Comparing (1) and (2)

`f'(x)=-(tan^(-1)x)` ---- (3)

Using power series expansion for
`tan^(-1)x`

`f'(x)=-tan^(-1)x=-\int {(1)/(1+x^(2)) \, dx`

`= -\int{ \sum\limits^( \infty)_(n=0) (-1)^(n)x^(2n)} \, dx`

`= -\sum{ \int\limits^( \infty)_(n=0) (-1)^(n)x^(2n)} \, dx`

`=-[c+\sum\limits^( \infty)_(n=0) (-1)^(n)(x^(2n+1))/(2n+1)]`

`=C+\sum\limits^( \infty)_(n=0) (-1)^(n+1)(x^(2n+1))/(2n+1)`

`=C-x+(x^(3))/(3)-(x^(5))/(5)+(x^(7))/(7)+.....`

as

`tan^(-1)(0)=0 \implies C=0`

Hence,

`\int{(-1)/(1+x^(2)) \, dx =-x+(x^(3))/(3)-(x^(5))/(5)+(x^(7))/(7)+.....`