Thursday

August 21, 2014

August 21, 2014

Posted by **nicholas** on Saturday, February 17, 2007 at 1:05pm.

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+1-1)/(2x+1) =

1-1/(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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