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

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

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)]

(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)]

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

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

Of course! I'd be happy to help you with that integral.

To evaluate the integral ∫(e^x+1)/(e^x +x) dx from 0 to 1, we'll use the method of definite integration.

Step 1: Start by expanding the numerator. We have (e^x+1), which can be written as e^x + 1.

Step 2: Now we rewrite the integral as ∫(e^x + 1)/(e^x + x) dx.

Step 3: Next, we'll perform partial fraction decomposition. We want to express the integrand as the sum of simpler fractions. To do this, we need to find the values of A and B in the form:

(e^x + 1)/(e^x + x) = A/(e^x) + B/(x)

To find the values of A and B, we'll multiply the entire equation by the denominator (e^x + x):

(e^x + 1) = A(x) + B(e^x + x)

Expanding the equation, we get:

e^x + 1 = A(x) + Be^x + Bx

Matching the coefficients on both sides of the equation, we can equate the coefficients of e^x, x, and constants separately.

e^x coefficient: 1 = B
Constant coefficient: 1 = A + B

Substituting B = 1 into the second equation, we get:

1 = A + 1
A = 0

So, we have A = 0 and B = 1.

Now we can rewrite the integrand as:

(e^x + 1)/(e^x + x) = 0/(e^x) + 1/(x)

Step 4: Now we can split the integral into two separate integrals. The first integral will be 0/(e^x), which is equal to 0 and can be ignored. So, we only need to evaluate the integral of 1/(x) from 0 to 1.

∫[0 to 1] 1/(x) dx = ln(x) |[0 to 1]

Step 5: Evaluating the integral, we have:

ln(1) - ln(0)

However, ln(0) is undefined, so we must take the limit as x approaches 0:

ln(1) - lim(x→0) ln(x)

Since ln(1) = 0, we only need to evaluate the limit:

lim(x→0) ln(x)

Step 6: Evaluating the limit:

lim(x→0) ln(x) = -∞

So, the value of the integral from 0 to 1 of (e^x+1)/(e^x + x) dx is -∞.

Please let me know if I can help you with anything else.