Evaluate the indefinite integral of (ln(x))/(x + xln(x)) dx by using variable substitution. Show all steps of your work.
First, see you problem posted
Wednesday, February 29, 2012 at 8:43am
for a correction to my answer there.
∫lnx / (x + xlnx) dx
You sure there's no typo here? If the problem were
∫lnx / (-x + xlnx) dx
it would be an example of that other integration problem, since
d/dx (-x + xlnx) = -1 + lnx + x/x = lnx
and letting g(x) = lnx we would have
∫dg/g = ln(g)
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using your problem as written,
∫lnx / (x + xlnx) dx
u = lnx
du = dx/x
and we have
∫lnx / (1 + lnx) * dx/x
= ∫u/(1+u) du
now let v = 1+u and we have
∫ (v-1)/v dv = ∫(1 - 1/v) dv
= v - ln(v)
1+u - ln(1+u)
= 1 + lnx - ln(1+lnx) + C
or, absorbing the 1,
lnx - ln(1+lnx) + C
To evaluate the indefinite integral, we will use variable substitution. We will substitute u for ln(x). Let's go through the steps:
Step 1: Let u = ln(x). This implies that du/dx = 1/x.
Step 2: To find dx in terms of u, we can rearrange the equation in Step 1 and solve for x: x = e^u.
Step 3: Substitute u and dx in terms of u into the integral:
∫(ln(x))/(x + xln(x)) dx = ∫(u)/(e^u + e^u*u) * (1)/(e^u) du.
Simplifying the integral, we have:
∫(u)/(e^u + e^u*u) * (1)/(e^u) du
= ∫(u)/(e^u + ue^u) du
Step 4: Now we have a new integral with u as the variable. We can further simplify this expression by factoring out e^u from the denominator:
∫(u)/(e^u(1 + u)) du
Step 5: Next, we apply partial fraction decomposition. We split the denominator into two terms:
∫A/e^u du + ∫B/(1 + u) du
Step 6: To solve for A and B, we set up the following equation using the original denominator:
(u) = A(1 + u) + Be^u
Expanding and equating coefficients, we get:
u = A + Au + Be^u
To find A and B, we equate the coefficients of like terms:
1 = A
1 = A + B
Solving these equations, we find that A = 1 and B = 0.
Step 7: Substituting the values of A and B back into the original integral, we get:
∫(u)/(e^u(1 + u)) du = ∫1/e^u du + ∫0/(1 + u) du
= ∫e^(-u) du + 0
= -e^(-u) + C
Step 8: Replacing u with ln(x), we get the final result:
-1/x + C
Therefore, the indefinite integral of (ln(x))/(x + xln(x)) dx is -1/x + C, where C is the constant of integration.