help!
∫((ln x^2)/x)) dx
u= ln x^2
du= 2 ln x dx (?)
Not sure how to set this one up?
let u = ln(x^2) or 2lnx
then du/dx = 2/x
du = (2/x)dx
let dv = (1/x) dx
dv/dx = 1/x
v = lnx
then
∫ u dv = uv - ∫v du
∫2lnx(1/x) dx = 2lnx(lnx) - ∫lnx(2/x)dx
2 ∫2lnx / x dx = 2 (lnx)^2
∫(ln x^2)/x dx = (lnx)^2
proof:
d (lnx)^2) /dx = 2(lnx)(1/x) = ln (x^2) / x
Thank you!
To solve the integral ∫((ln x^2)/x)) dx, you're on the right track by using u-substitution. Let's go through the step-by-step process together:
1. Start by expressing the integral in terms of u using the substitution u = ln(x^2).
∫((ln x^2)/x)) dx = ∫(u/x) dx
2. Now we need to find dx in terms of du to fully substitute:
Differentiate both sides of the equation u = ln(x^2) with respect to x:
du/dx = 2(1/x)
Multiply both sides by dx:
du = 2(1/x) dx
We can rearrange this equation to solve for dx:
dx = (x/2) du
3. Substitute u and dx in terms of du into the original integral:
∫(u/x) dx = ∫(u/(x/2)du = 2∫(u/(2x)) du
4. Notice that we can simplify further by factoring out the constant 2:
2∫(u/(2x)) du = 2 * (1/2) * ∫(u/x) du = ∫(u/x) du
5. Now, the integration becomes simpler:
∫(u/x) du = ∫(u * (1/x)) du = ∫(u/x) du
6. Since the integrand is ∫(u/x) du, we can now integrate with respect to u while treating x as a constant:
∫(u/x) du = (∫u du/x)
7. Evaluate the integral:
(∫u du/x) = (1/x) * ∫u du
The integral of u with respect to u is (u^2/2):
(1/x) * ∫u du = (1/x) * (u^2/2)
8. Substitute back u = ln(x^2):
(1/x) * (u^2/2) = (1/x) * (ln(x^2)^2/2)
9. Simplify the expression:
(ln(x^2)^2/2x) is the final result of the integral ∫((ln x^2)/x)) dx.
Therefore, the solution to ∫((ln x^2)/x)) dx is (ln(x^2)^2/2x).