Integral of x sq/(1+x sq) dx
I am not sure how to do this, am i supposed to do partial fractions to break it up? Or is it just some simple substitution?
If I see the derivative of the base sitting out front either directly of as a multiple, I usually can do the integral intuitively
I know it has to be (?)(1 + x^2)^(3/2)
I then visulaize differentiating that, and adjust the coefficient accordingly.
so it should be (1/3)(1 + x^2)^(3/2)
Your teacher probably wants you to do it by substitution
let u = (1 + x^2)
du/dx = 2x
dx = du/2x
so Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
= (1/3)u^(3/2)
= (1/3)(1 + x^2)^(3/2)
But how did you get
Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
du= 2xdx, from there i got,
integral xsq/(1+x) dx
= (1/2) integral[ (sqrt (1-u))/u ]du
Did i do anything wrong?
To solve the integral ∫(x^2/(1+x^2)) dx, we can actually use a simple substitution technique. Here's how you can approach it:
1. Start by letting u = 1 + x^2. This substitution simplifies the integral by introducing a new variable.
Therefore, du/dx = 2x, which implies dx = (1/2x) du.
2. Now substitute the values of x and dx in the integral based on the substitution made. We have:
∫(x^2/(1+x^2)) dx = ∫((x^2) / (u)) (1/2x) du
= (1/2) ∫(x/ u) du
3. Simplify the integrand by canceling out the 'x' terms:
∫(x/ u) du = (1/2) ∫(1/u) du
4. The integral of 1/u is a well-known result: ln|u|. So, we have:
∫(1/u) du = ln|u|
5. Now substitute back u = 1 + x^2:
∫(x^2/(1+x^2)) dx = (1/2) ln|u|
= (1/2) ln|1 + x^2|
Thus, the solution to the integral ∫(x^2/(1+x^2)) dx is (1/2) ln|1 + x^2|.