Anybody know how to integrate ln(2e^x)?

Answer 1

first, you take ln(2e^x) and change it into
ln(2)+ln(e^x)
that becomes
ln(2) + x
constant
then integrate to get
ln(2)x + x^2/2 + C

Answer 2

Integrate ln(2e^x).

ln(2e^x) = ln 2 + ln e^x = ln 2 + x

∫ln(2e^x)dx = ∫{ln 2 + x)dx = (ln 2)x + x²/2 + C

Answer 3

ln(2e^x)=ln2 + ln e^2=ln 2 +x
∫(ln 2 +x=x ln 2 + 1/2 x^2 c=.5x^2+(ln 2) x + c

Answer 4

ln(2*e^x) = ln2 + x so the integral is ln2*x +x^2/2(In is log base"e"

if In means integral it is 2e^x

Answer 5

ln(2e^x) =

ln(2) + ln(e^x) =

ln(2) + x ln e =

ln(2) + x

Integrating would give you

x ln(2) + 1/2 x² + C

Answer 6

use integration by parts

ln(2e^x)

ln(2e^x)*1
uv-interate(v(du/dx))

integration of ln(2e^x)*1 =xln(2e^x)-integration ofx

=xln(2e^x)-x^2/2+c