Find using substitution:
∫ 1/[x(1+lnx)^2] dx
I'm not sure what substitution to use.
maybe something like
let z = ln x
dz /dx = 1/x
dx = x dz
∫ 1/[x(1+lnx)^2] dx
∫ dz/[ 1+z]^2
Don't worry, I've got it. I used the substitution u = 1 + lnx and got the answer
To find the integral ∫ 1/[x(1+lnx)^2] dx, you can use the substitution method. Here's the step-by-step process:
Step 1: Choose a substitution.
In this case, let's choose u = 1 + ln(x). This substitution allows us to simplify the expression under the integral sign.
Step 2: Find the derivative of the substitution (du/dx) and solve for dx.
Differentiating both sides of the substitution equation u = 1 + ln(x) with respect to x, we get du/dx = 1/x.
Rearranging this equation, we have dx = du / (1/x) = x du.
Step 3: Substitute the variables and simplify the expression.
Using the substitution u = 1 + ln(x) and dx = x du, rewrite the integral as ∫ (1/[x(1+lnx)^2]) dx = ∫ (1/[u^2]) (x du).
Since x = e^(u-1), we can further simplify the expression as ∫ (1/[u^2]) (e^(u-1) du).
Step 4: Evaluate the integral.
The integral ∫ (1/[u^2]) (e^(u-1) du) can be evaluated easily.
Using the property that the integral of (1/u^2) du is -1/u (i.e., ∫ (1/[u^2]) du = -1/u), we have:
∫ (1/[u^2]) (e^(u-1) du) = -e^(u-1)/u + C
Step 5: Convert back to the original variable.
Substituting u = 1 + ln(x) back into the equation, we get:
- e^(1 + ln(x) - 1)/(1 + ln(x)) + C
Simplifying further, we have:
- e^ln(x)/(1 + ln(x)) + C
- x/(1 + ln(x)) + C
Therefore, the result of the integral is ∫ 1/[x(1+lnx)^2] dx = - x/(1 + ln(x)) + C.