integration by parts
posted by nicholas .
s integral
s ln (2x+1)dx ?
= ln(2x+1)x  s x d( ln (2x+1))
= ln(2x+1)x s x [(2x+1)'/ (2x+1)] dx
= ln(2x+1)x s x [(2)/ (2x+1)] ?... then i'm confused...
"ln(2x+1)x s x [(2)/ (2x+1)] ?... then i'm confused..."
x [(2)/ (2x+1)] =
2x/(2x+1) =
(2x+11)/(2x+1) =
11/(2x+1)
B.t.w. first substituting 2x + 1 = u before doing partial integration simplifies things.
Also differentiation w.r.t. a parameter is often simpler than partial integration. E.g. you can calculate the integral of Ln(x) as follows. You consider the integral:
Integral x^a dx = 1/(a+1) x^(a+1)
Differentiate both sides w.r.t. the parameter a. Note that:
x^a = Exp[a Ln(x)] >
d x^a/d a = Ln(x) Exp[a Ln(x)] =
Ln(x) x^a
Integral x^a Ln(x)dx =
1/(a+1)^2 x^(a+1)
+ 1/(a+1) x^(a+1)Ln(x)
If you substitute a = 0 in here you get the result:
Integral Ln(x)dx = x Ln(x)  x
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