# How do I find the antiderivative of #f(x)=1/(e^(2x)-9)^(1/2)#?

Try this:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the antiderivative of ( f(x) = \frac{1}{\sqrt{e^{2x} - 9}} ), you can use the substitution method. Let ( u = e^{x} ), then ( du = e^{x} dx ). Rearranging, we have ( dx = \frac{du}{u} ). Substitute ( u ) and ( du ) into the integral:

[ \int \frac{1}{\sqrt{u^2 - 9}} \frac{du}{u} ]

This becomes:

[ \int \frac{1}{\sqrt{u^2 - 9}} \cdot \frac{du}{u} ]

Now, make another substitution: ( v = \frac{u}{3} ), then ( u = 3v ) and ( du = 3dv ). Rearranging, we have ( dv = \frac{du}{3} ). Substitute ( v ) and ( dv ) into the integral:

[ \int \frac{1}{\sqrt{(3v)^2 - 9}} \cdot \frac{du}{3} ]

[ = \int \frac{1}{\sqrt{9(v^2 - 1)}} \cdot \frac{1}{3} dv ]

[ = \frac{1}{3} \int \frac{1}{\sqrt{v^2 - 1}} dv ]

Now, this integral can be recognized as the integral of ( \sec(\theta) ) with respect to ( \theta ). The antiderivative of ( \sec(\theta) ) is ( \ln|\sec(\theta) + \tan(\theta)| + C ).

So, the antiderivative of ( f(x) ) is:

[ \frac{1}{3} \ln\left| \sec(\theta) + \tan(\theta) \right| + C ]

Where ( \theta ) can be found using the substitution ( v = \sec(\theta) ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7