Question

Please help me... and show steps please :)

integral from 0 to 1 of (e^x+1)/(e^x +x) dx

Answer 1

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)]

Answer 2

(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)]

Answer 3

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

Answer 4

Let C be the portion of the curve y =8ãx between (1,8) and (25, 40). FindçC 2yds.