What is the antiderivative of #(1)/(1+x^2)#?

Answer 1
The antiderivative of #1/(1+x^2)# is the integral
#int 1/(1+x^2)dx# which is equivalent to
#int 1/(1+x^2)dx=arctanx+C#
where #arctanx# is the inverse of the trigonometric function
#tanx# and C is the integration constant.
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Answer 2

# = arctan(x) + c #

Let

#color(blue)(x = tantheta #
#=> color(red)(dx = sec^2 theta d theta) " By the use of the quotient rule..." #
#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 -= 1 #
#=> sin^2 x / cos^2 x + cos^2x/cos^2x -= 1/cos^2 x #
#=> tan^2x + 1 -= sec^2 x #
#=> int sec^2 theta / sec^2 theta d theta #
#=> int 1 d theta #
#=> theta + c#
If #x = tan theta => arctanx = theta #

Replace once more in...

#color(blue)( arctan(x ) + c #
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Answer 3

The antiderivative of ( \frac{1}{1+x^2} ) with respect to ( x ) is ( \arctan(x) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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