How do you integrate $\int ln (2 x + 1) d x$ $intln(2x+1)dx$ ?

Question

How do you integrate $\int ln (2 x + 1) d x$ $intln(2x+1)dx$ ?

1 Answer

Impact of this question

1543 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-07T00:41:05

I would set $2 x + 1 = t$ $2x+1=t$
$2 d x = d t$ $2dx=dt$
$\int ln (t) \frac{d t}{2} =$ $intln(t)dt/2=$ by parts:
$= \frac{1}{2} [t ln (t) - \int t \cdot \frac{1}{t} d t] =$ $=1/2[tln(t)-intt*1/tdt]=$
$= \frac{1}{2} [t ln (t) - t + c]$ $=1/2[tln(t)-t+c]$
but $t = 2 x + 1$ $t=2x+1$
$= \frac{1}{2} [(2 x + 1) ln | 2 x + 1 | - (2 x + 1) + c]$ $=1/2[(2x+1)ln|2x+1|-(2x+1)+c]$

Science

Math

Humanities

... and beyond

How do you integrate $\int ln (2 x + 1) d x$ $intln(2x+1)dx$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫ln(2x+1)dxintln(2x+1)dx?

1 Answer

Related questions

Impact of this question

How do you integrate $\int ln (2 x + 1) d x$ $intln(2x+1)dx$ ?