s- integral
s ln (2x+1)dx ?
= ln(2x+1)x - s x d( ln (2x+1))
= ln(2x+1)x- s x [(2x+1)'/ (2x+1)] dx
= ln(2x+1)x- s x [(2)/ (2x+1)] dx
= ln(2x+1)x- s (2x/2x+1) dx
= ln(2x+1)x- s [(2x+1-1)/(2x+1)] ... i'm confused about why there is a minus 1 after the integral
The minus sign in this case appears because of the application of a rule called integration by parts. The rule is derived from the product rule for differentiation, which states that (u*v)' = u'v + uv'.
The integration by parts formula is given as follows:
∫(u*dv) = uv - ∫(v*du)
In your case, we let u = ln(2x+1) and dv = dx. Taking the derivatives, we have du = (1/(2x+1))*d(2x+1) = dx and v = x.
Now, we can apply the integration by parts formula to the integral:
∫(ln(2x+1)*dx) = ln(2x+1)*x - ∫(x*(1/(2x+1))*d(2x+1))
Simplifying further, we have:
∫(ln(2x+1)*dx) = ln(2x+1)*x - ∫(x*(2/(2x+1)) dx
The integral ∫(x*(2/(2x+1)) dx can be evaluated separately. In order to do that, we need to simplify the integrand:
∫(x*(2/(2x+1)) dx = ∫(2x/(2x+1)) dx = ∫((2x+1-1)/(2x+1)) dx
Expanding the numerator, we have:
∫((2x+1-1)/(2x+1)) dx = ∫(1 - 1/(2x+1)) dx
Now we can evaluate the integral of 1 and the integral of 1/(2x+1) separately. The integral of 1 is simply x, and the integral of 1/(2x+1) can be found by using a substitution or a u-substitution.
So, to summarize:
∫(ln(2x+1)*dx) = ln(2x+1)*x - ∫(x*(2/(2x+1)) dx = ln(2x+1)*x - (x - ln|2x+1|) + C
where C is the constant of integration.