What's the integral of....?

Question

What's the integral of 4/(t^2+1) dt with bounds of 0 and 1 and how do you find it?

Answer 1

I = ∫ 4 / ( t ² + 1 ² ) dt
I = 4 tan ^(-1) t-------- between limits 0 and 1
I = 4 tan^(-1) (1)
I = 4 ( π / 4 )
I = π

Answer 2

Note that the integral of 1/(1+t^2) dt = arctan(t) + C. Since this is a definite integral, we can do away with the C, evaluate the indefinite integral at the limits of integration, and just multiply by 4: 4*arctan(1) - 4*arctan(0). This is 4*(Pi/4), or Pi. (arctan(1) is Pi/4 because tan(Pi/4) = 1, and arctan(0) = 0 because tan(0) = 0.)

Answer 3

Integral of 4 / (t^2 + 1 ) dt = 4 tan inverse t

Now applying the bounds, it is

4 tan inverse 1 - 4 tan inverse 0 = 4 ( pi/4 ) = pi

Answer 4

The integral of dx/(c²+x²) is, per my CRC handbook,
1/c arctan(x/c).

In this case, c=1 and you have a multiplier of 4, so the integral is:
4 arctan t

You find the definite integral by subtracting the value at t=1 from the value at t=0.