Find the indicated integral

Question

∫x/(2x+1) ⅆx

Answer 1

Let u = 2x+1  Then, x = (u-1)/2  and dx = (1/2)du

 

So, ∫[x/(2x+1)]dx = ∫[(u-1)/(2u)](1/2)du

 

                          = ¼∫[1 - 1/u]du = ¼[ u - ln lul ] + C₁

 

                                               = ¼[ (2x+1) - ln l 2x+1 l + C₁

 

                                               = ½x - ¼ln l 2x+1 l + C  (C = C₁+¼)  

Answer 2

Rewrite the integral so that it has no denominators.

 

∫x(2x + 1)^-1dx

 

 

Now we can use integration by parts.  We know that the formula using integration by parts is

 

∫udv = uv - ∫vdu

 

 

Let                   u = x                              dv = (2x + 1)^-1dx

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

 

 

Plugging in, we get

 

∫x(2x + 1)^-1dx = (1/2)xln(2x + 1) - (1/2)∫ln(2x + 1)dx

 

 

Next, we find the integral of   ln(2x + 1)dx  using substitution.

 

Let             r = 2x + 1

               dr = 2dx

               (1/2)dr = dx

 

 

∫ln(2x + 1)dx = (1/2)∫ln(r)dr

                     = (1/2)[rln(r) - r]

                     = (1/2)[(2x + 1)ln(2x + 1) - (2x + 1)]

 

Lets denote this part of the result B. 

 

So the final integral will be

 

∫x(2x + 1)^-1dx = (1/2)xln(2x + 1) - (1/2)B

 

 

Plug in that value of B into the final integral and simplify.

 

 

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