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General antiderivative of a sum of two standard forms

Find the general antiderivative of g(x)=11−x2+31+x2.g(x) = \frac{1}{\sqrt{1-x^2}} + \frac{3}{1+x^2}.

Hint

Recall the standard derivative facts ddx[arcsin⁡x]=11−x2\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}} and ddx[arctan⁡x]=11+x2\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}. Reverse each of these separately, then combine the results with a single arbitrary constant.

Step-by-step solution

The function g(x)g(x) is a sum of two terms, so we can antidifferentiate each term separately using linearity. For the first term, recall that ddx[arcsin⁡x]=11−x2\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}, so an antiderivative of 11−x2\frac{1}{\sqrt{1-x^2}} is arcsin⁡x\arcsin x.

For the second term, recall that ddx[arctan⁡x]=11+x2\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}, so an antiderivative of 31+x2\frac{3}{1+x^2} is 3arctan⁡x3\arctan x. Adding these pieces together and including a single arbitrary constant of integration gives the general antiderivative G(x)=arcsin⁡x+3arctan⁡x+CG(x) = \arcsin x + 3\arctan x + C.

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