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)] ?... then i'm confused...
"ln(2x+1)x- s x [(2)/ (2x+1)] ?... then i'm confused..."
x [(2)/ (2x+1)] =
2x/(2x+1) =
(2x+1-1)/(2x+1) =
1-1/(2x+1)
B.t.w. first substituting 2x + 1 = u before doing partial integration simplifies things.
Also differentiation w.r.t. a parameter is often simpler than partial integration. E.g. you can calculate the integral of Ln(x) as follows. You consider the integral:
Integral x^a dx = 1/(a+1) x^(a+1)
Differentiate both sides w.r.t. the parameter a. Note that:
x^a = Exp[a Ln(x)] --->
d x^a/d a = Ln(x) Exp[a Ln(x)] =
Ln(x) x^a
Integral x^a Ln(x)dx =
-1/(a+1)^2 x^(a+1)
+ 1/(a+1) x^(a+1)Ln(x)
If you substitute a = 0 in here you get the result:
Integral Ln(x)dx = x Ln(x) - x
To integrate the expression ∫ ln(2x+1) dx, you can use the technique of integration by parts.
Start by applying the formula for integration by parts:
∫ u dv = uv - ∫ v du
In this case, let u = ln(2x+1) and dv = dx.
Differentiate u to find du:
du = (1/(2x+1)) * 2 dx = 2/(2x+1) dx
Integrate dv to find v:
v = ∫ dx = x
Now plug in the values of u, du, v, and dv into the integration by parts formula:
∫ ln(2x+1) dx = uv - ∫ v du
= x ln(2x+1) - ∫ x * (2/(2x+1)) dx
= x ln(2x+1) - 2 ∫ x/(2x+1) dx
The next step is to simplify the remaining integral, which is ∫ x/(2x+1) dx. One way to do this is by using a substitution.
Let u = 2x+1, then du = 2 dx.
Rearrange this equation to solve for dx:
dx = (1/2) du
Substitute these values into the integral:
∫ x/(2x+1) dx = ∫ (1/2) * (x/u) du
= (1/2) ∫ (x/u) du
= (1/2) ∫ (1 - 1/u) du
= (1/2) ∫ (1 - 1/(2x+1)) du
= (1/2) (u - ln|2x+1|) + C
= (1/2) (2x+1 - ln|2x+1|) + C
Finally, substitute this result back into the previous equation:
∫ ln(2x+1) dx = x ln(2x+1) - 2((1/2) (2x+1 - ln|2x+1|) + C)
= x ln(2x+1) - (2x+1 - ln|2x+1|) + C
= x ln(2x+1) - 2x - 1 + ln|2x+1| + C
So the final result of the integral is:
∫ ln(2x+1) dx = x ln(2x+1) - 2x - 1 + ln|2x+1| + C