Respuesta :
Answer:
Required series is:
[tex]\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....[/tex]
Step-by-step explanation:
Given that
[tex]f'(x) = -\frac{1}{1 + x^{2}}[/tex] ---(1)
We know that:
[tex]\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}[/tex] ---(2)
Comparing (1) and (2)
[tex]f'(x)=-(tan^{-1}x)[/tex] ---- (3)
Using power series expansion for [tex]tan^{-1}x[/tex]
[tex]f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx [/tex]
[tex]= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx [/tex]
[tex]= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx [/tex]
[tex]=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}][/tex]
[tex]=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}[/tex]
[tex]=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....[/tex]
as
[tex]tan^{-1}(0)=0 \implies C=0[/tex]
Hence,
[tex]\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....[/tex]
