Asked by Simone
Evaluate the indefinite integral of (ln(x))/(x + xln(x)) dx by using variable substitution. Show all steps of your work.
Answers
Answered by
Steve
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)
----------------
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
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)
----------------
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
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