# integral help

• Nov 29th 2009, 11:26 PM
amma0913
integral help
How do I find the integral of

F(x) = the integral of 1/1+t^2 dt from 0 to x + the integral of 1/1+t^2 dt from 0 to 1/x?

and how do I determine that F(x) is constant on (-infinity,0) and constant on (0,infinity)?

Any help appreciated!

Thanks
• Nov 30th 2009, 01:55 AM
mr fantastic
Quote:

Originally Posted by amma0913
How do I find the integral of

F(x) = the integral of 1/1+t^2 dt from 0 to x + the integral of 1/1+t^2 dt from 0 to 1/x?

and how do I determine that F(x) is constant on (-infinity,0) and constant on (0,infinity)?

Any help appreciated!

Thanks

You should know that $\int \frac{1}{1 + t^2} \, dt = \tan^{-1} (t) + C$. Apply this standard result and simplify (using a right triangle might be helpful ....)
• Nov 30th 2009, 02:01 AM
tonio
Quote:

Originally Posted by amma0913
How do I find the integral of

F(x) = the integral of 1/1+t^2 dt from 0 to x + the integral of 1/1+t^2 dt from 0 to 1/x?

and how do I determine that F(x) is constant on (-infinity,0) and constant on (0,infinity)?

Any help appreciated!

Thanks

$F(x)=\int\limits_0^x\frac{dt}{1+t^2}+\int\limits_0 ^{1\slash x}\frac{dt}{1+t^2}=\left[\arctan t\right]_0^x+\left[\arctan t\right]_0^{1\slash x}=\arctan x+\arctan\left(1\slash x\right)$.

To determine that this function is a constant in some interval just differentiate it...what should you get?

Tonio
• Nov 30th 2009, 03:41 AM
HallsofIvy
Since (F+ G)'= F+ G', you could also show this function is a constant by differentiating each of the integrals separately and adding. By the "Fundamental Theorem of Calculus" the derivative of the first integral is just $\frac{1}{1+ x^2}$. By the "Fundamental Theorem of Calculus", together with the chain rule, the derivative of the second integral is $\frac{1}{1+ \frac{1}{x^2}}\left(\frac{-1}{x^2}\right)= -\frac{1}{1+ x^2}$. Those clearly add to 0, for all x (except x= 0 where $\frac{1}{1+ \frac{1}{x^2}}$ is not defined), so the function is a constant.