Come trovi l'integrale di $x {(ln x)}^{3} d x$ $x (ln x) ^ 3 dx$ ?

Questo ovviamente Integrazione per parti.

Permettere:
$u = {ln}^{3} x$ $u = ln^3x$
$d u = \frac{3 {ln}^{2} x}{x} d x$ $du = (3ln^2x)/xdx$
$d v = x d x$ $dv = xdx$
$v = \frac{x^{2}}{2}$ $v = x^2/2$

$u v - \int v d u$ $uv - intvdu$

$= x^2/2ln^3x - int x^cancel(2)/2 * (3ln^2x)/cancel(x)dx$

$= (x^2ln^3x)/2 - 3/2int xln^2xdx$

Ripetere:
$u = ln^2x$
$du = (2lnx)/xdx$
$dv = xdx$
$v = x^2/2$

$= (x^2ln^3x)/2 - 3/2(int xln^2xdx)$

$= (x^2ln^3x)/2 - 3/2((x^2ln^2x)/2 - int x^cancel(2)/cancel(2) * (cancel(2)lnx)/cancel(x) dx)$

$= (x^2ln^3x)/2 - 3/2((x^2ln^2x)/2 - int xlnx dx)$

e ripeti ancora:
$u = lnx$
$du = 1/xdx$
$dv = xdx$
$v = x^2/2$

$= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2(int xlnx dx)$

$= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2((x^2lnx)/2 - int x^cancel(2)/2*1/cancel(x)dx)$

$= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2((x^2lnx)/2 - int x/2dx)$

$= 1/2x^2ln^3x - 3/4x^2ln^2x + 3/4x^2lnx - 3/4int xdx$

$= 1/2x^2ln^3x - 3/4x^2ln^2x + 3/4x^2lnx - 3/8x^2$

$= color(blue)(1/8x^2(4ln^3x - 6ln^2x + 6lnx - 3) + C)$

Come trovi l'integrale di x (ln x) ^ 3 dx x(lnx)3dxx (ln x) ^ 3 dx ?

Lascia un commento Annulla risposta

Come trovi l'integrale di $x {(ln x)}^{3} d x$ $x (ln x) ^ 3 dx$ ?