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How do you integrate #int4/(x^2 + 9)dx#?

Answer 1

#4/3arctan(x/3)+C#

Let #x=3t#, then: #dx=3dt,# #int4/(x^2+9)dx=int4/(9t^2+9)3dt=int4/9(t^2+1)3dt=# #=4/3intdt/(t^2+1)=4/3arctant+C=4/3arctan(x/3)+C#

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Answer 2

Adam Adkins

To integrate ( \frac{4}{x^2 + 9} ) with respect to ( x ), you can use the substitution method. Let ( u = x/3 ), then ( du = dx/3 ). Thus, ( dx = 3du ). Substituting ( u = x/3 ) and ( dx = 3du ), the integral becomes ( \int \frac{4}{9(u^2 + 1)} \cdot 3 du ). Simplifying, we have ( \frac{4}{3} \int \frac{1}{u^2 + 1} du ). This integrates to ( \frac{4}{3} \arctan(u) + C ), where ( C ) is the constant of integration. Finally, substituting back ( u = x/3 ), we get ( \frac{4}{3} \arctan(x/3) + C ) as the final result.

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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.