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)] =
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)] =
Integral x^a Ln(x)dx =
+ 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
Answer this Question
calc - evaluate the integral: y lny dy i know it's integration by parts but i ...
calculus-integration - integrate -2/xln^4(x)...plz help me..give me an idea on ...
Calc 121 - How do you integrate using substitution: the integral from 1 to 3 of...
calculus - Use integration by parts to evaluate the integral of x*sec^2(3x). My ...
calc asap! - can you help me get started on this integral by parts? 4 S sqrt(t) ...
integration by parts - s- integral s ln (2x+1)dx ? = ln(2x+1)x - s x d( ln (2x+1...
Calculus - Use integration by parts to find the integral. Round the answer to ...
integration by part - s- integral s (p^5) lnp dp ? = (lnp)[(p^6)/(6)] - s (i'm ...
Calc - Use integration by parts to evaluate the integral. ∫ from 1 to 4 of...
math - integration by parts - Integration of x^2 ln(x^3+2)