What is the antiderivative of $\frac{1}{1 + x^{2}}$ $(1)/(1+x^2)$ ?

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What is the antiderivative of $\frac{1}{1 + x^{2}}$ $(1)/(1+x^2)$ ?

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Answer 1 · 2016-01-23T01:25:58

The antiderivative of $\frac{1}{1 + x^{2}}$ $1/(1+x^2)$ is the integral

$\int \frac{1}{1 + x^{2}} d x$ $int 1/(1+x^2)dx$ which is equivalent to

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

where $arctan x$ $arctanx$ is the inverse of the trigonometric function

$tan x$ $tanx$ and C is the integration constant.

Answer 2 · 2018-02-03T11:05:05

$= arctan (x) + c$ $= arctan(x) + c$

Explanation:

Let

$x = tan θ$ $color(blue)(x = tantheta$

$\Rightarrow d x = {sec}^{2} θ d θ By the use of the quotient rule...$ $=> color(red)(dx = sec^2 theta d theta) " By the use of the quotient rule..."$

$\int \frac{1}{1 + x^{2}} d x = \int \frac{1}{1 + {(tan θ)}^{2}} \cdot {sec}^{2} θ d θ$ $int 1/(1+color(blue)(x)^2 ) color(red)(dx) = int 1/(1+color(blue)((tantheta))^2 ) * color(red)(sec^2 theta d theta$

We know ${sin}^{2} x + {cos}^{2} x \equiv 1$ $sin^2 x + cos^2 x -= 1$

$\Rightarrow \frac{{sin}^{2} x}{{cos}^{2} x} + \frac{{cos}^{2} x}{{cos}^{2} x} \equiv \frac{1}{{cos}^{2} x}$ $=> sin^2 x / cos^2 x + cos^2x/cos^2x -= 1/cos^2 x$

$\Rightarrow {tan}^{2} x + 1 \equiv {sec}^{2} x$ $=> tan^2x + 1 -= sec^2 x$

$\Rightarrow \int \frac{{sec}^{2} θ}{{sec}^{2} θ} d θ$ $=> int sec^2 theta / sec^2 theta d theta$

$\Rightarrow \int 1 d θ$ $=> int 1 d theta$

$\Rightarrow θ + c$ $=> theta + c$

If $x = tan θ \Rightarrow arctan x = θ$ $x = tan theta => arctanx = theta$

Substitute back in...

$arctan (x) + c$ $color(blue)( arctan(x ) + c$

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What is the antiderivative of $\frac{1}{1 + x^{2}}$ $(1)/(1+x^2)$ ?

2 Answers

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