Answer:
Required series is:
![\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....](https://tex.z-dn.net/?f=%5Cint%7B%5Cfrac%7B-1%7D%7B1%2Bx%5E%7B2%7D%7D%20%5C%2C%20dx%20%3D-x%2B%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D-%5Cfrac%7Bx%5E%7B5%7D%7D%7B5%7D%2B%5Cfrac%7Bx%5E%7B7%7D%7D%7B7%7D%2B.....)
Step-by-step explanation:
Given that
---(1)
We know that:
---(2)
Comparing (1) and (2)
---- (3)
Using power series expansion for ![tan^{-1}x](https://tex.z-dn.net/?f=tan%5E%7B-1%7Dx)
![= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx](https://tex.z-dn.net/?f=%3D%20-%5Cint%7B%20%5Csum%5Climits%5E%7B%20%5Cinfty%7D_%7Bn%3D0%7D%20%28-1%29%5E%7Bn%7Dx%5E%7B2n%7D%7D%20%5C%2C%20dx%20)
![= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx](https://tex.z-dn.net/?f=%3D%20-%5Csum%7B%20%5Cint%5Climits%5E%7B%20%5Cinfty%7D_%7Bn%3D0%7D%20%28-1%29%5E%7Bn%7Dx%5E%7B2n%7D%7D%20%5C%2C%20dx%20)
![=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]](https://tex.z-dn.net/?f=%3D-%5Bc%2B%5Csum%5Climits%5E%7B%20%5Cinfty%7D_%7Bn%3D0%7D%20%28-1%29%5E%7Bn%7D%5Cfrac%7Bx%5E%7B2n%2B1%7D%7D%7B2n%2B1%7D%5D)
![=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}](https://tex.z-dn.net/?f=%3DC%2B%5Csum%5Climits%5E%7B%20%5Cinfty%7D_%7Bn%3D0%7D%20%28-1%29%5E%7Bn%2B1%7D%5Cfrac%7Bx%5E%7B2n%2B1%7D%7D%7B2n%2B1%7D)
![=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....](https://tex.z-dn.net/?f=%3DC-x%2B%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D-%5Cfrac%7Bx%5E%7B5%7D%7D%7B5%7D%2B%5Cfrac%7Bx%5E%7B7%7D%7D%7B7%7D%2B.....)
as
![tan^{-1}(0)=0 \implies C=0](https://tex.z-dn.net/?f=tan%5E%7B-1%7D%280%29%3D0%20%5Cimplies%20C%3D0)
Hence,
![\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....](https://tex.z-dn.net/?f=%5Cint%7B%5Cfrac%7B-1%7D%7B1%2Bx%5E%7B2%7D%7D%20%5C%2C%20dx%20%3D-x%2B%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D-%5Cfrac%7Bx%5E%7B5%7D%7D%7B5%7D%2B%5Cfrac%7Bx%5E%7B7%7D%7D%7B7%7D%2B.....)
Answer:
c. i took the test
Step-by-step explanation:
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Answer:
7i
Step-by-step explanation:
negative numbers have no real square roots. However there are "imaginary numbers which can be used as the sqrt of negative numbers