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How do you integrate #(1/(16+x^2))#?

Answer 1

Ezekiel Alexander

#int 1/(16+x^2)dx=1/4arc tan (x/4)+C#.

We know that #int1/(a^2+x^2)dx=1/aarc tan(x/2)+C#.

#:. int 1/(16+x^2)dx=1/4arc tan (x/4)+C#.

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Answer 2

Charles Anctil

To integrate ( \frac{1}{16+x^2} ), you can use the following method:

Make a substitution ( x = 4 \tan(\theta) ) or ( x = 4 \cot(\theta) ).

Then, ( dx = 4 \sec^2(\theta) , d\theta ).

Substitute these values into the integral, and you'll get:

[ \int \frac{1}{16+x^2} , dx = \int \frac{1}{16+16\tan^2(\theta)} \cdot 4\sec^2(\theta) , d\theta ]

This simplifies to:

[ = \int \frac{4\sec^2(\theta)}{16(1+\tan^2(\theta))} , d\theta ]

[ = \int \frac{4\sec^2(\theta)}{16\sec^2(\theta)} , d\theta ]

[ = \frac{1}{4} \int , d\theta ]

[ = \frac{\theta}{4} + C ]

Finally, revert back to the original variable ( x ) using the substitution ( x = 4 \tan(\theta) ).

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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.