anonymous
  • anonymous
Can someone explain how can I integrate \int { { x }^{ 2 } } { tan }^{ -1 }xdx
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Draw it
anonymous
  • anonymous
\int { { x }^{ 2 } } { tan }^{ -1 }xdx\[\int\limits { { x }^{ 2 } } { \tan }^{ -1 }xdx\]
anonymous
  • anonymous
Use integration by parts

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
sounds about right. Thanks! :)
anonymous
  • anonymous
fun! integration by parts gives you u=atan(x) and du=1/(1 + x^2) dv=x^2 and v=(x^3)/3 thus the first integration gives you: ((x^3)*atan(x))/3 - (1/3)int((x^3)/((1+x^2))dx messy... but the second integration can be simplified in a very creative way by adding +x-x to the numerator. by doing this, you can make a term of x^3 into x(x^2 + 1), which cancels with the denominator and leaves you integrating just x and a second term of -x/(1+x^2) which if you set u=1+x^2 and du=2x you can integrate u^-1 and end up with something that looks like -(1/2)ln|1+x^2|. distribute the -1/3 outside the integral sign back in, and you should get your answer. (check my work, but I think that's right). oh, and don't forget the +C for indefinite integrals! I would write the solution, but I think it would make less sense typed than it does on the paper I just wrote it on. hopefully you are in cal2 or none of this will make a lick of sense. good luck!
anonymous
  • anonymous
Should I get a medal too? :D
anonymous
  • anonymous
lol... First off.. Dude you write SOOOO fast! You gave me a whole essay worth of answer lol.. and yeah for sure.. if you could teach me how to do that :)
anonymous
  • anonymous
Where's my medal xD Grr..well here it goes *takes breather* We know that arc tan x is differentiable u = arctan x du/dx = 1/1+x^2 x^2dx = dv so v = x^3/3 now int udv = uv - int vdu/dx dx = x^3(arctan x) - int (x^3/3(1+x^2) dx) let 1+ x^ 2 = t x^2 = (t-1) 2xdx = - dt so we get x^3/1(1+x^2) dx = x^2xdx/(1+x^2)/3 now you can proceed as = (t-1) (-dt/6)/t = -1/6(dt- dt/t) integrating we get -1/6 t + 1/6ln(t) as t >0 substiture the value and get the result
anonymous
  • anonymous
you're tooo adorable. and i ment how do i give you a metal? Meaning where do i go in order to give it to you specifically? lol
anonymous
  • anonymous
Click the "best response" next to my answers! xD
anonymous
  • anonymous
that was easy! :)
anonymous
  • anonymous
Honestly thanks a WHOLE LOT!! :D
anonymous
  • anonymous
Anytime <3
anonymous
  • anonymous
\[Let I=\int\limits x ^{2} \tan^{-1} x dx\] \[put \tan^{-1} x =y,x=\tan y,dx=\sec ^{2}y dy\] \[I=\int\limits \tan ^{2}y \left( y \right)\sec ^{2}y dy=\int\limits y \left( \tan ^{2}y\right)\left( \sec ^{2}y dy \right)\] integrate by parts \[I=y \frac{ \tan ^{3}y }{ 3 }-\int\limits 1.\frac{ \tan ^{3}y }{3 } dy+c\] \[I=y \frac{ \tan ^{3}y }{ 3} -\frac{ 1 }{ 3}\int\limits \tan y .\tan ^{2}y dy+c\] \[I=y \frac{ \tan ^{3}y }{ 3 } -\frac{ 1 }{ 3 } \int\limits \tan y \left( \sec ^{2}y -1 \right)dy+c\] \[I=y \frac{ \tan ^{3}y }{ 3 }-\frac{ 1 }{ 3 }\int\limits \tan y \sec ^{2}y dy+\frac{ 1 }{ 3 }\int\limits \frac{- \sin y }{\cos y }dy+c\] \[I=y \frac{ \tan ^{3}y }{ 3 }-\frac{ 1 }{ 3 }\frac{ \tan ^{2}y }{ 2 }+\frac{ 1 }{ 3 }\ln \left| \cos y \right|\] |dw:1370723765885:dw| \[\cos y=\frac{ 1 }{\sqrt{1+x ^{2}} }\] substitute the values of cos y & tan y
anonymous
  • anonymous
correction write -1/3 instead of 1/3 write also +c at the end.

Looking for something else?

Not the answer you are looking for? Search for more explanations.