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.