How do you find the indefinite integral of #int 1/(3x^2+3)dx#?

Answer 1

#int 1/(3x^2 + 3)dx = 1/3arctanx + C#

Factor out a #3# in the denominator.
#int 1/(3(x^2 + 1))#
#1/3int 1/(x^2 + 1)#

We now have a known integral.

#1/3arctanx + C#

Hopefully this helps!

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Answer 2

To find the indefinite integral of ( \int \frac{1}{3x^2 + 3} , dx ), you can use a trigonometric substitution. Let ( u = x\sqrt{3} ). Then ( du = \sqrt{3} , dx ).

After substitution, the integral becomes ( \int \frac{1}{u^2 + 1} , du ), which is a standard integral with respect to ( u ). This integral evaluates to ( \arctan(u) + C ), where ( C ) is the constant of integration.

Substituting back ( u = x\sqrt{3} ), the final result is ( \arctan(x\sqrt{3}) + C ).

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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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