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.

Anyone know how to do this integral?

Mathematics
See more answers at brainly.com
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

\[\int\limits \frac{ x ^{2} +x+1}{ (x ^{2}+1)^{2} }dx\]
Play around with the fraction.
Maybe partial fraction decomposition

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

did you try a trig sub i think that would work just fine
you think i can try sin?
I think you can try tan
the hint was the x^2+1 on bottom
tan^2(theta)+1=sec^2(theta)
would the set up be integral x^2+x+1/sex^2 theta
well you have to replace all the x's and the dx
and also you are leaving off the square on bottom
\[x=\tan(\theta) => dx=\sec^2(\theta) d \theta \] Replace all the x's with tan(theta) Replace the dx with sec^2(theta) d theta
is it integral tan^2theta+tan theta+1/sec^2 theta d (theta)
If I think what you wrote is what I think then yes. You mean the following I assume: integral of (tan^2(theta)+tan(theta)+1)/sec^2(theta) d(theta)
\[\int\limits_{}^{}\frac{\tan^2(\theta)}{\sec^2(\theta)} d \theta +\int\limits_{}^{}\frac{\tan(\theta)}{\sec^2(\theta)} d \theta +\int\limits_{}^{} \frac{1}{\sec^2(\theta) }d \theta \]
Look at them three separately
\[\int\limits_{}^{}(\frac{ x^2+1 }{ (x^2+1)^2 } +\frac{ x }{ (x^2+1)^2 })*dx\]
\[\int\limits_{}^{}\frac{ dx }{ x^2+1 } +\int\limits_{}^{}\frac{ x }{ (x^2+1)^2 }\]
first one is standard formula
how can i figure out the answer?
\[\int\limits_{}^{}\frac{ x }{ (x^2+1)^2 } *dx\] for this x^2+1=t 2xdx=dt xdx=dt/2
\[\int\limits_{}^{} \frac{ dt }{ 2*t^2 }\]
\[\int\limits_{}^{}\frac{ dx }{ x^2+1 } +\int\limits_{}^{}\frac{ dt }{ 2t^2 }\]
can u solve it @mony01
is it sec^2 (theta)/2(x^2+1)^2
I do like gorv's way. But either way is fine. gorv's is simpler though.
actual its less calculative...both ways anns will be same
how can i solve the rest of it?
What question on what part do you have?
\[\int\limits \frac{ dx }{ x ^{2}+1}+\int\limits \frac{ dt }{ 2t ^{2} }\]
the first integral is just remembering that it is acran(x) the second one 1/2 is a constant multiple and you should know how to integrate t^(-2) at this point in calculus
is the answer arctan(x)-1/2(x^2+1)+C
Yes that's right you mean arctan(x)- 1/[2(x^2+1)] +C good job

Not the answer you are looking for?

Search for more explanations.

Ask your own question