How do you integrate #1/(x^2 + 9)#?

Answer 1

#1/3arctan(x/3)+C#

We will try to put this in the form of the arctangent integral:

#int1/(u^2+1)du=arctan(u)+C#

So here, we see that:

#int1/(x^2+9)dx=int1/(9(x^2/9+1))dx=1/9int1/((x/3)^2+1)dx#
Let #u=x/3#, implying that #du=1/3dx#:
#=1/3int(1/3)/((x/3)^2+1)dx=1/3int1/(u^2+1)du=1/3arctan(x/3)+C#
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Answer 2
#intdx/(x^2+9)#
Let #x=3tantheta#, implying that #dx=3sec^2thetad theta#.
#=int(3sec^2thetad theta)/(9tan^2theta+9)#
Since #tan^2theta+1=sec^2theta#:
#=1/3intsec^2theta/sec^2thetad theta=1/3intd theta=1/3theta+C#
Reversing the original substitution #x=3tantheta#:
#=1/3tan^-1(x/3)+C#
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Answer 3

To integrate 1/(x^2 + 9), you can use the formula for the integral of a rational function involving a quadratic term in the denominator. This can be done by applying a trigonometric substitution. The integral simplifies to (1/3) * arctan(x/3) + 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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