Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__37.gif

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

i did this with u substituion it did not work
i am getting something like ln2/2

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

It looks to me more an atan(2x)
how
i took u=1+4t^2
du/8t=dt
The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)
|dw:1358025284718:dw|
So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)
use the equation button
The result is not ln(u)
ok i should remember that formula right?
I suggest you use y=2t
Yes, it's on the first page of standard integrals. Check it out.
ok
\( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)
@mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }+c\]
Right, so a=1/2.
what is x then
\[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c\]
x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)
and a=1/2
can i do \[\frac{ 1 }{ 1 }\tan^{-1} \frac{ 2t }{ 1}\]
and t=1/2
t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the left-hand side is exactly your question. All you need to do after that is to evaluate the right-hand side.

Not the answer you are looking for?

Search for more explanations.

Ask your own question