How do you integrate $\int \frac{1}{1 + x^{2}} + \int \frac{x}{1 + x^{2}} d x$ $int1/(1+x^2) + int x/(1+x^2) dx$ ?

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How do you integrate $\int \frac{1}{1 + x^{2}} + \int \frac{x}{1 + x^{2}} d x$ $int1/(1+x^2) + int x/(1+x^2) dx$ ?

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Answer 1 · 2018-04-01T17:29:25

$\int \frac{d x}{1 + x^{2}} + \int \frac{x}{1 + x^{2}} = arctan x + ln \sqrt{1 + x^{2}} + C$ $int dx/(1+x^2) + int x/(1+x^2) = arctanx + ln sqrt (1+x^2) +C$

Explanation:

The first integral is a well known anti-derivative:

$\int \frac{d x}{1 + x^{2}} = arctan x + C$ $int dx/(1+x^2) = arctanx +C$

For the second we can substitute:

$x^{2} = u$ $x^2 = u$

$2 x d x = d u$ $2xdx = du$

so:

$\int \frac{x}{1 + x^{2}} = \frac{1}{2} \int \frac{d u}{1 + u} = ln \sqrt{| 1 + u |} + C$ $int x/(1+x^2) = 1/2 int (du)/(1+u) = ln sqrt abs (1+u)+C$

then:

$\int \frac{d x}{1 + x^{2}} + \int \frac{x}{1 + x^{2}} = arctan x + ln \sqrt{1 + x^{2}} + C$ $int dx/(1+x^2) + int x/(1+x^2) = arctanx + ln sqrt (1+x^2) +C$

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How do you integrate $\int \frac{1}{1 + x^{2}} + \int \frac{x}{1 + x^{2}} d x$ $int1/(1+x^2) + int x/(1+x^2) dx$ ?

1 Answer

Explanation:

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