Integral Problem: u-substitution for int [1/(2x + 1)] dx using u = 2x + 1

[Vulcan]

Trying to find the following integral, any pointers?

---

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{2x\, +\, 1}\, dx\)

. . . . .Let u = 2x + 1; then \(\displaystyle \,\dfrac{du}{dx}\, =\, 2,\, \) so \(\displaystyle \, du\, =\, 2\, dx\)

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{u}\, dx\)

. . . . .I know it needs to be:

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{u}\, du\)

. . . . .but not sure how to get there.

---

Thanks in advance for any assistance.

 

[Deleted member 4993]

Trying to find the following integral, any pointers?

---

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{2x\, +\, 1}\, dx\)

. . . . .Let u = 2x + 1; then \(\displaystyle \,\dfrac{du}{dx}\, =\, 2,\, \) so \(\displaystyle \, du\, =\, 2\, dx\)

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{u}\, dx\). . .Incorrect

. . . . .I know it needs to be:

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{u}\, du\)

. . . . .but not sure how to get there.

---

Thanks in advance for any assistance.

\(\displaystyle \displaystyle {\int \dfrac{1}{2x+1}dx \ = \ \int \dfrac{1}{u}\frac{du}{2} \ = \ \frac{1}{2}\int \dfrac{1}{u}du}\)

 

[Vulcan]

\(\displaystyle \displaystyle {\int \dfrac{1}{2x+1}dx \ = \ \int \dfrac{1}{u}\frac{du}{2} \ = \ \frac{1}{2}\int \dfrac{1}{u}du}\)

I see where I went wrong now.

Many thanks