Verified answer

It might help to do a substitution t = 2x + 1, dt = 2dx first.

∫ ln(2x + 1) dx = ∫ ½ ln(t) dt

Now, for parts, take u = ln(t), du = 1/t dt and dv = dt so v = t

∫ ½ ln(t) dt = ½[ t ln(t) - ∫ t(1/t) dt] = ½ t ln(t) - ½ ∫ dt = ½ t ln(t) - ½ t + C.

Going back to x

∫ ln(2x + 1) dx = ½(2x + 1) ln(2x + 1) - ½(2x + 1) + C.

This can be written slightly differently. Since ½(2x + 1) = x + ½, you can combine the ½ in this sum with the constant C of integration and write

∫ ln(2x + 1) dx = ½(2x + 1) ln(2x + 1) - x + C.

By parts:

u = ln (2x + 1), du = 2/(2x+1), dv = dx, v = x

uv - Integral v du = x ln(2x+1) - Integral 2x/(2x+1) dx

Integral of 2x/(2x + 1) dx = Integral of 1 - 1/(2x+1) dx = x - 1/2 ln (2x + 1)

Answer: x ln (2x +1) - x + 1/2 ln (2x+1) + C

∫ ln(2x+1) is integrated by parts.

∫ ln(2x+1) = ln(2x+1) ∫1.dx - ∫[(2x+1)'. ∫1.dx].dx

= ln(2x+1)(x) - ∫[(2+0).(x)]dx

= ln(2x+1)(x) - 2 ∫xdx

= ln(2x+1)(x) - 2(x^2/2) + c

= ln(2x+1)(x) - x^2 + c

Thats the answer.whenever u get to integrate LOG or INVERSE TRIGNOMETRIC integral always consider the 1st function and integrate them by parts

i will coach each step as you asked. 2·? x·ln(2x) dx u = ln(2x) du = [a million/x] dx dv = x dx v = ½·x² ? u dv = uv - ? v du 2·? x·ln(2x) dx = 2·[ ½·x²·ln(2x) - ? ½·x²·[a million/x] dx 2·? x·ln(2x) dx = 2·[ ½·x²·ln(2x) - ½·? x dx] 2·? x·ln(2x) dx = 2·[ ½·x²·ln(2x) - ¼·x²] + C 2·? x·ln(2x) dx = x²·ln(2x) - ½·x² + C

Let u = 2x + 1, du = 2dx, dx = du/2

Now integrate by parts.

v = ln(u), dv = dx/x

w = u, dw = du

∫ln(2x + 1)dx = vw – 1/2∫u du = 1/2[(2x + 1)ln(2x + 1) – 2x + 1] + K

use the tabular method and put ln(2x+1) as your U and DX as you DV

u would derive to 1/[2(2x+1)] and DX would integrate to x then you would get

xln(2x+1)- ∫x/[2(2x+1)]dx

you can probably solve from there but i'll put the rest down

xln(2x+1)-1/2∫x/(2x+1)

then u=2x+1

1/2du=dx

x=(u-1)/2

xln(2x+1)-1/8∫(u-1)/u du

xln(2x+1)-1/8∫1-1/u du

then intigrate

xln(2x+1)-1/8u+1/8ln|u|+c

xln(2x+1)-1/8(2x+1)+1/8ln|2x+1|+C

t = 2x + 1

dt = 2dx

ln(2x + 1) * dx =

ln(t) * (1/2) * dt =

(1/2) * ln(t) * dt

int((1/2) * ln(t) * dt) =

(1/2) * int(ln(t) * dt)

u = ln(t)

du = dt/t

dv = dt

v = t

(1/2) * int(u * dv) =

(1/2) * (uv - int(v * du)) =

(1/2) * (t * ln(t) - int(t * dt / t)) =

(1/2) * (t * ln(t) - int(dt)) =

(1/2) * (t * ln(t) - t) + C =

(1/2) * (t) * (ln(t) - 1) + C

t = 2x + 1

(1/2) * (2x + 1) * (ln(2x + 1) - 1) + C

Let u = ln(2x+1) so du = 2dx/(2x+!)

Let dv = dx so v = x

uv - ∫vdu = xln(2x+1) - ∫2x/(2x+1)dx

uv - ∫vdu = xln(2x+1) - [∫(2x+1 - 1)/(2x+1)dx]

uv - ∫vdu = xln(2x+1) - ∫(2x+1)/(2x+1)dx + ∫1/(2x+1)dx

uv - ∫vdu = xln(2x+1) - x + ½ln(2x+1)

uv - ∫vdu = (x - uv - ∫vdu = (x+½)ln(2x+1) - x