Home » Ask & Discuss » Mathematics. » Integral Calculus « Back to Discussion

Integral Calculus

Blazing goIITian

Joined:	12 Apr 2008
Post:	2717

5 Nov 2008 22:40:04 IST

0 People liked this

535

Evaluate:

None

$\int \cot^{-1}(x^2+x+1)\ dx$

Please help me..

Share this article on:

Comments (7)

Mirka

Blazing goIITian

Joined: 13 Aug 2008

Posts: 1313

5 Nov 2008 22:44:54 IST

Like 2 people liked this

This is cot inverse of ( 1 + x(1+x) ) / 1+x - x )

= arc cot ( 1+x ) - arc cot ( x )

then u can use that Integral of cot inverse x is
x. arc cot x + 1/2 ln ( 1 + x^2 ) + c

Ankit

Blazing goIITian

Joined: 17 Oct 2007

Posts: 1659

6 Nov 2008 21:00:44 IST

Like 0 people liked this

http://integrals.wolfram.com/index.jsp?expr=arccot(x^2+%2Bx%2B1)&random=false

Ashutosh Sharma

Blazing goIITian

Joined: 29 Sep 2008

Posts: 628

6 Nov 2008 21:06:10 IST

Like 0 people liked this

Gr8 integrator...

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

6 Nov 2008 22:00:18 IST

Like 0 people liked this

OK here is my solution:

$I=\cot^{-1 }(x^2+x+1)\ dx$

$=\int \tan^{-1 }\left(\frac 1{x^2+x+1}\right)\ dx$

$=\int \{\tan^{-1 }(x+1)-\tan^{-1}(x)\}dx$

$=\int \tan^{-1}(x+1)\ dx-\int \tan^{-1}(x)\ dx$

Applying inteegration by parts, we get:

$=\{ \tan^{-1}(x+1)\}x-\int x\frac 1{1+(x+1)^2}\ dx-\left\{(\tan^{-1} x)x-\int x \frac 1{1+x^2}\ dx\right\}$

$=x\{\tan^{-1}(x+1)-\tan^{-1}(x)\}+\frac 1{2}\log(1+x^2)-I_a$

Let $I_a=\int \frac x{1+(1+x)^2}\ dx$ and put x+1=t, dx=dt

$=\int \frac {t-1}{1+t^2}\ dt=\int \frac t{1+t^2}\ dt-\int \frac 1{1+t^2}\ dt$

$=\frac 1{2}\log|1+t^2|-\tan^{-1}(t)$

$I_a=\frac 1{2}\log|1+(x+1)^2|-\tan^{-1}(1+x)$

Do the rest and you will get the answer... a bit long.

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

7 Nov 2008 16:19:08 IST

Like 0 people liked this

rudra..so tht was a challenge !!! ...
so why help was req. in first post..

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

7 Nov 2008 20:24:24 IST

Like 0 people liked this

" rudra..so tht was a challenge !!! ...
so why help was req. in first post.."

No it was not..I kept trying this for the whole day in the school and i got the wrong answer and you call it a challenge. lol

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

7 Nov 2008 21:34:44 IST

Like 0 people liked this

:) :) shud have mentioned it..really cretes a bad impression..!! lol

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team