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

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Dang i wish i could , good luck (i’m bad at math)
Someone its due in 30 min i dont know how to do pls
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Required series is:

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

Step-by-step explanation:

Given that

                           f'(x) = -\frac{1}{1 + x^{2}} ---(1)

We know that:

                  \frac{d}{dx}(tan^{-1}x)=\frac{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 {\frac{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}\frac{x^{2n+1}}{2n+1}]

=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}

=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

as

                 tan^{-1}(0)=0 \implies C=0

Hence,

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....



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