Question

Evaluate (copied):
\[ \int\cot^{-1} (x^2+x+1)dx \]

Answer 1

\[\int\cot^{-1}\left(x^2+x+1\right)dx\\
\int\cot^{-1}\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)dx\]
\[u=x+\frac{1}{2}\\
du=dx\]
\[\int\cot^{-1}\left(u^2+\frac{3}{4}\right)du\]
Integrating by parts, let
\[\begin{matrix}f=\cot^{-1}\left(u^2+\frac{3}{4}\right)& &dg=du\\
df=-\frac{2u}{\left(u^2+\frac{3}{4}\right)^2+1}du& &g=u\end{matrix}\]
\[u\cot^{-1}\left(u^2+\frac{3}{4}\right)+\int\frac{2u^2}{\left(u^2+\frac{3}{4}\right)^2+1}du\]
I'm not sure how to approach this next integral, though... I was thinking a sub like
\[u=\sqrt t, \text{ then maybe }\\t+\frac{3}{4}=\tan y,\]
but it doesn't look like it'll all work out well. Wolframalpha suggests partial fractions, but I don't see how that would work.

Answer 2

try integration by parts ...

Answer 3

You mean on the newest integral, right?

Answer 4

no .. on the original one. and try to use the property of arctan ... this is freaking tricky, otherwise it would get complicated.

Answer 5

Ah, that makes it much simpler.

Answer 6

\[\cot ^{-1}x=\tan ^{-1}\frac {1}{x}\]\[\int \tan^{-1}\frac{1}{x}dx\]

Answer 7

not sure is this right substitution

Answer 8

Yep, that's how it looks it will start off.

Answer 9

yep ... that's one!! there is also another property of arctan ... related to addition.

Answer 10

\[\int \tan^{-1}\frac{1}{x^2+x+1}dx\]

Answer 11

\[\tan^2 x+1=\cot ^2x\]???

Answer 12

That's not the right identity.

Answer 13

tan^2x+1=sec^2x

Answer 14

No not that one.
\[ \arctan a + \arctan b = ?\]

Answer 15

pi/2??

Answer 16

a and b related how

Answer 17

no ..
\[ \arctan a + \arctan b = \arctan \left( \frac{a+b}{1-ab}\right)\]

Answer 18

see if you can apply there.

Answer 19

from \[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A*\tan B}\]

Answer 20

why dont we take f=tan 1/x^2+x+1 g=1\[I=x(\tan^{-1}\frac{1}{x^2+x+1})-\int\limits \frac{1}{1+(x^2+x+1)^2}*x\]

Answer 21

g'=1 g=x

Answer 22

yeah, the identity follows from that formula ... but, see if can make it work ...
that's terrible ...

Answer 23

you mean my attempt above is terrible or the identity

Answer 24

the attempt ... i warned in the my second post.

Answer 25

how will you break down this ...
\[ \int\limits \frac{x}{1+(x^2+x+1)^2} dx\]

Answer 26

\[ \frac{1}{1+(x^2+x+1)^2}+\frac{x-1}{1+(x^2+x+1)^2}\] can it works

Answer 27

is this partial fract

Answer 28

yeah ... how will do you the first one?

Answer 29

also it's -1 ... 1+x, but the results are same.

Answer 30

first one???

Answer 31

this does not look nice.