2 answers:
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<h2>Explanation:</h2>
We have the function:
The vertical asymptote can be found when the denominator equal zero, therefore:
So the vertical asymptote occur at
You have one mistake which occurs when you integrate 
. The antiderivative of this is not in terms of 
. Instead, letting 
(or 
, if you want to bother with more signs) gives 
, making the indefinite integral equality
and then compute the definite integral from there.
Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting
. Then 
, and the integral becomes
Another way to do this is to notice that the integrand's denominator can be factorized.
So,
There are no discontinuities to worry about since you're integrate over
, so you can proceed with integrating straightaway.
Just goes to show there's often more than one way to skin a cat...
and then compute the definite integral from there.
Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting
Another way to do this is to notice that the integrand's denominator can be factorized.
So,
There are no discontinuities to worry about since you're integrate over
Just goes to show there's often more than one way to skin a cat...
Answer:
x = 30
Step-by-step explanation:
i hope it help, sorry if its wrong
