Question

Can someone explain how can I integrate
\int { { x }^{ 2 } } { tan }^{ -1 }xdx

Answer 1

Draw it

Answer 2

\int { { x }^{ 2 } } { tan }^{ -1 }xdx\[\int\limits { { x }^{ 2 } } { \tan }^{ -1 }xdx\]

Answer 3

Use integration by parts

Answer 4

sounds about right. Thanks! :)

Answer 5

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!

Answer 6

Should I get a medal too? :D

Answer 7

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 :)

Answer 8

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

Answer 9

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

Answer 10

Click the "best response" next to my answers! xD

Answer 11

that was easy! :)

Answer 12

Honestly thanks a WHOLE LOT!! :D

Answer 13

Anytime <3

Answer 14

\[Let I=\int\limits x ^{2} \tan^{-1} x dx\]
\[put \tan^{-1} x =y,x=\tan y,dx=\sec ^{2}y dy\]
\[I=\int\limits \tan ^{2}y \left( y \right)\sec ^{2}y dy=\int\limits y \left( \tan ^{2}y\right)\left( \sec ^{2}y dy \right)\]
integrate by parts
\[I=y \frac{ \tan ^{3}y }{ 3 }-\int\limits 1.\frac{ \tan ^{3}y }{3 } dy+c\]
\[I=y \frac{ \tan ^{3}y }{ 3} -\frac{ 1 }{ 3}\int\limits \tan y .\tan ^{2}y dy+c\]
\[I=y \frac{ \tan ^{3}y }{ 3 } -\frac{ 1 }{ 3 } \int\limits \tan y \left( \sec ^{2}y -1 \right)dy+c\]
\[I=y \frac{ \tan ^{3}y }{ 3 }-\frac{ 1 }{ 3 }\int\limits \tan y \sec ^{2}y dy+\frac{ 1 }{ 3 }\int\limits \frac{- \sin y }{\cos y }dy+c\]
\[I=y \frac{ \tan ^{3}y }{ 3 }-\frac{ 1 }{ 3 }\frac{ \tan ^{2}y }{ 2 }+\frac{ 1 }{ 3 }\ln \left| \cos y \right|\]


\[\cos y=\frac{ 1 }{\sqrt{1+x ^{2}} }\]
substitute the values of cos y & tan y

Answer 15

correction write -1/3 instead of 1/3
write also +c at the end.