Question regarding integral of ln(2x+1) dx

Question

∫ln(2x+1) dx; when I solve for this using u-sub with u = 2x+1, I get 1/2((2x+1)ln(2x+1) - (2x+1)) + C. However, when I solve for it using integration by parts, I get 1/2((2x+1)ln(2x+1)) - x + C. Various sources give both answers interchangeably, whether it be integration calculators or solution manuals, but both answers aren't the same; when I set them equal to each other, it comes out as 0 = -1, which is obviously not true. Which one is the right answer, and why? Is it because 1/2 + C can be simplified to just C? And whenever a number is added to C, can it just be simplified to C?

Answer 1

Yes, you can always lump more constants into the +C for indefinite integrals. It can be deceptive but the +C for one integration method isn't necessarily equal to the +C for another, even if you start off with the same integrand.

 

Especially for logarithms, an answer like ln[(x²+1)/5]+C would be equivalent to ln[x²+1]+C: ln[(x²+1)/5] = ln(x²+1) - ln(5), and ln(5) is a sneaky constant that could be grouped into C.

 

A similar thing happens with ∫sin(x) cos(x)dx depending on which function you use for a u-substitution. In that case you can show the solutions are the same by using sin²(x)+cos²(x)=1. You could also integrate using the double angle identity sin(x)cos(x)=(1/2)sin(2x) to obtain a third result, which again can be equated to the other two.

Answer 2

You need integration by parts for this one:

 

   u = ln(2x+1)     dv = dx

 

   du = [2 / (2x+1)]]dx   v = x

 

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

 

                                                                        Let w = 2x+1  then x = (w-1)/2 and dw = 2dx

 

                                           = xln(2x+1) - (1/2)∫[(w-1)/w]dw

 

                                           = xln(2x+1) - (1/2)∫[1 - 1/w]dw

 

                                           = xln(2x+1) - (1/2)w + (1/2)ln(2x+1) + C₁

 

                                           = xln(2x+1) + (1/2)ln(2x+1) - (1/2)(2x+1) + C₁

 

                                           = xln(2x+1) + (1/2)ln(2x+1) - x + C   (C = C₁ - 1/2)

 

This answer is equivalent to yours!