Question

what is the integration of tan^-1(1/(x^2-x+1))

Answer 1

I think integration by parts is the way to go here. Let
\[\begin{matrix}
u=\tan^{-1}\left(\frac{1}{x^2-x+1}\right)&dv=dx\\
du=\cdots~~~~~~~~~~~~~~~~~~~~~~~~~&v=x
\end{matrix}\]
Completing the square in the denominator and making a substitution there might make finding \(du\) easier.

Answer 2

Hmm yah I think sith has the right idea here.

\[\large= x \arctan\left(\frac{1}{x^2-x+1}\right)-\int\limits x\frac{\left(\dfrac{1}{x^2-x+1}\right)'}{1+\left(\dfrac{1}{x^2-x+1}\right)^2}dx\]

That new integral looks like a doozy to clean up.. but it's probably the right way to go.

Answer 3

Complete the square of the denominator and it should look familiar.

Answer 4

\[\large x^2-x+1 \qquad = \qquad x^2-x+\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)^2+1\]\[\large =\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\]

@NoelGreco , it should look familiar? D:
Hmm I'm not quite seeing it...

Answer 5

\[\int\limits x\frac{\left(\dfrac{1}{x^2-x+1}\right)'}{1+\left(\dfrac{1}{x^2-x+1}\right)^2}dx=\int\limits -x\frac{\left(\dfrac{2x-1}{(x^2-x+1)^2}\right)}{1+\left(\dfrac{1}{x^2-x+1}\right)^2}dx\]\[\int \frac{-x(2x-1)}{1+(x^2-x+1)^2} dx\]how to do it from here?

Answer 6

i can not understand the given process........sorry

Answer 7

is there any one to help.............

Answer 8

Unless there is some really important reason to do otherwise, this looks like a candidate for numerical integration. I'd like to see the ENTIRE original problem statement.

Answer 9

ok the original problem statement is to,,,intigrate....tan^-1(1/x^2-x+1)

Answer 10

answer brought to you by Matlab. It doesn't show steps tho.

arctan(x - 1) + ln(x^2 - 2*x + 2)/2 - ln(x^2 + 1)/2 + x*arctan(1/(x^2 - x + 1))

Answer 11

\[\large \int\limits \frac{-x}{1+(x^2-x+1)^2} (2x-1)dx\]

\(\large u=x^2-x+1\)
\(\large du=(2x-1)dx\)

\[\large \int\limits \frac{-x}{1+u^2}du\]

Hmmmmm......

Answer 12

wht is meant by arctan?@chemENGINEER

Answer 13

how have u strted?????@zepdrix

Answer 14

hey one thing @zepdrix how come we have x left in the integral if we are integrating w.r.t u?i mean it will then be treated as a constnt isn't it?

Answer 15

No not constant, it needs to be dealt with.
So there's an issue there :P that's why i stopped.
Kinda stuck.

Answer 16

@shaown I'm not sure if the start process was correct, so I'm not sure if it's worth explaining in detail.. hmm

Answer 17

Are you familiar with `Integration by Parts` Shaown? :o

Answer 18

\[ \arctan \left( \frac{1}{x^2 - x + 1} \right) = \arctan\left( \frac{x - (x-1)}{1 + x(x-1)}\right ) = \arctan x - \arctan(x-1) \]
From here, use Integration by parts.

Answer 19

neat as usual @experimentX