Integral of 1/(2x+1): different methods give different results??

[Acko93]

Integral of 1/(2x+1)

Way to solve this is to sub t=2x+1 and solution is (1/2)*ln(2x+1) +c

But if i pull 1/2 in front of integral and sub t=x+1/2 solution is (1/2)*ln(x+1/2)+ c

Is these two solutions same and constants are different?

 

[stapel]

Integral of 1/(2x+1)

Way to solve this is to sub t=2x+1 and solution is (1/2)*ln(2x+1) +c

But if i pull 1/2 in front of integral and sub t=x+1/2 solution is (1/2)*ln(x+1/2)+ c

Is these two solutions same and constants are different?

Yes. Set the two results equal to each other, using different constants:

. . . . .\(\displaystyle \dfrac{1}{2}\, \ln(2x\, +\, 1)\, +\, C\, =\, \dfrac{1}{2}\, \ln\left(x\, +\, \dfrac{1}{2}\right)\, +\, D\)

Multiply through by 2, move the logs onto one side of the equation and the constants to the other, combine the two logs into one and simplify, and then solve for one of the constants in terms of the other.

 

[JeffM]

Integral of 1/(2x+1)

Way to solve this is to sub t=2x+1 and solution is (1/2)*ln(2x+1) +c

But if i pull 1/2 in front of integral and sub t=x+1/2 solution is (1/2)*ln(x+1/2)+ c

Is these two solutions same and constants are different?

This is a good question.

The terminology can be confusing. When we say "plus a constant" with respect to an indefinite integral, we really mean "plus some number." What you did, writing down c to stand for a number in both expressions, leads to the question "how can these two different expressions be equal."

Stapel's solution of using C and D for the two different expressions is clever. But C and D are related. Although C can be any number, D is fixed once C is specified, and vice versa.

\(\displaystyle \dfrac{1}{2} * ln (2x + 1) + C = \dfrac{1}{2} * ln \left (x + \dfrac{1}{2} \right ) + D \implies\)

\(\displaystyle \dfrac{1}{2} * ln (2x + 1) + C = \dfrac{1}{2} * ln \left (\dfrac{2}{2} * \left \{ x + \dfrac{1}{2} \right \} \right ) + D \implies\)

\(\displaystyle \dfrac{1}{2} * ln (2x + 1) + C = \dfrac{1}{2} * ln \left ( \dfrac{1}{2} \{2x + 1\} \right ) + D \implies\)

\(\displaystyle \dfrac{1}{2} * ln (2x + 1) + C = \dfrac{1}{2} * \left \{ ln \left ( \dfrac{1}{2} \right ) + ln(2x + 1) \right \} + D \implies\)

\(\displaystyle \dfrac{1}{2} * ln (2x + 1) + C = \dfrac{1}{2} * ln (2x + 1) + \dfrac{1}{2} * ln \left ( \dfrac{1}{2} \right ) + D \implies\)

\(\displaystyle C = \dfrac{1}{2} * ln \left ( \dfrac{1}{2} \right ) + D.\)

C and D are not independent but are different numbers.

 

[lookagain]

Integral of 1/(2x+1)

Way to solve this is to sub t=2x+1 and solution is (1/2)*ln(2x+1) +c

The natural log will take the absolute value bars in conjunction here:

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

 

[sinx]

good question

Integral of 1/(2x+1)

sub t=2x+1 and solution is (1/2)*ln(2x+1) +c

sub t=x+1/2 solution is (1/2)*ln(x+1/2)+ c

It is easier to see the result is the same if you integrate and apply definite limits, say 1 to x, (where x is upper limit, 1 is lower limit).
then y=(¹/₂) ln[(2x+1)/3];
both times.

good question.