Can you help me with these integrals?

Question

I've been racking my brain today trying to figure out the hang of these things. I can do some of them but these few got me stumped... Id really appreciate it if someone could help me out on this.

This one I think I might have gotten the right answer or close...
[-1,1] x/1+x^2 dx
I got 0.

These two I can't seem to get.

[0,1] 4/1+x^2 dx
&
[0,4] (2x+3) (square root of) x+4 dx

Thanks

Answer 1

Ok, let's go on :

[-1,1] x/1+x^2 dx

integrate( x / 1 + x^2)dx

x = tan(t) >>>>> dx = sec^2(t)

Replacing :

integrate((tan(t) / sec^2(t))*sec^2(t))dt

integrate(tan(t))dt

ln(cost) = ln(cos(arctan(x))

x = -1 to 1

ln(cos(pi/4)) - ln(cos(-pi/4) = ln(cos(pi/4) - ln(cos(pi/4)) = 0

2) [0,1] 4/1+x^2 dx

integrate( 4 / 1 + x^2)dx

4*arctan(x) ; x goes from 0 to 1

4*arctan(0) - 4*arctan(1) = 0 - 4*pi/4 = -pi

3) [0,4] (2x+3) (square root of) x+4 dx

integrate((2x+3)*sqrt(x+4))dx

integrate(2x*sqrt(x+4)dx + 3*integrate(sqrt(x+4))dx

(x+4)^5/2 / 5/2 - 4*(x+4)^5/2 / 3/2

2/5*(x+4)^5/2 - 8/3*(x+4)^5/2 + 2*(x+4)^3/2

Now you can replace x from 0 to 4.

Hope that helps

Answer 2

I'll do the first two and get you started on the last one.
1). You don't have to do any work on this one.
Look: x/(1+x²) is an odd function because
f(-x) = -f(x). The integral of any odd function
from -a to a is zero. So you are right. The answer is 0.

2. This one is standard. An antiderivative of 4/(1+x²)
is 4 arctan x. Evaluating this from 0 to 1 gives 4 arctan 1
which equals 4*π/4 = π.
Let's see why the derivative of arctan x is 1/(1+x²).
Let y = arctan x. Then x = tan y.
Taking derivatives and using the chain rule, we get
1 = sec² y dy/dx = (1+ tan² y) dy/dx.
So dy/dx = 1/ (1 + tan² y) = 1/(1+x²).

3. Here we have a so- called linear irrationality.
We can handle it by substitution.
Let u = x+4, du = dx, x = u-4.
Substituting and changing limits, we get
∫ (4..8) (2u-5)√u du.
Now just use the power rule to finish the problem.
Hope that helps!