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How do you find the integral of #dx/(1+x^2)#?

Answer 1

Isaiah Allen

#tan^-1 x +C#

Let #x = tan theta#, so that #dx=sec^2 theta d theta#

#int (1dx)/(1+x^2) # = #int (sec^2 theta d theta)/ sec^2 theta#

#int d theta = theta +C#

= #tan^-1 x +C#

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Answer 2

William Abdalla

To find the integral of ( \frac{dx}{1+x^2} ), you can use the arctangent integral formula, which states that:

[ \int \frac{1}{1+x^2} , dx = \arctan(x) + C ]

where ( C ) is the constant of integration. Therefore, the integral of ( \frac{dx}{1+x^2} ) is:

[ \int \frac{dx}{1+x^2} = \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.