split it up:
e^x dx/(e^x+1) + dx/(e^x+1)
in general:
integral e^ax dx/(b+ce^ax) = (1/ac)ln(b+ce^ax)
and
integral dx/(b+ce^ax) = 1/(ab)[ax-ln(b+ce^ax)]
Please help me... and show steps please :)
integral from 0 to 1 of (e^x+1)/(e^x +x) dx
4 answers
(e^x+1)/(e^x +x) =
f'(x)/f(x)
with f(x) = e^x +x
Integral of f'(x)/f(x) dx
can be computed by putting
f(x) = y
then
f'(x)dx = dy
So, the integral can be written as:
Integral of dy/y
Integration limits:
if x = 0, then y = f(x) = 1
if x = 1, then y = f(1) = 1 + exp(1)
So, the integral is:
Log[1 + exp(1)]
f'(x)/f(x)
with f(x) = e^x +x
Integral of f'(x)/f(x) dx
can be computed by putting
f(x) = y
then
f'(x)dx = dy
So, the integral can be written as:
Integral of dy/y
Integration limits:
if x = 0, then y = f(x) = 1
if x = 1, then y = f(1) = 1 + exp(1)
So, the integral is:
Log[1 + exp(1)]
since the derivative of ln(u) = u'/u
I recognized that the derivative of e^x +x is e^x + 1, thus having the exact pattern as noted above
so ∫(e^x + 1)/(e^x + x) dx = ln(e^x + x) + c
I recognized that the derivative of e^x +x is e^x + 1, thus having the exact pattern as noted above
so ∫(e^x + 1)/(e^x + x) dx = ln(e^x + x) + c
Let C be the portion of the curve y =8�ãx between (1,8) and (25, 40). Find�çC 2yds.