Question

Find d/dx integral from 1 to x^2 ln(t)dt

Answer 1

\[\frac{ d }{ dx } \int\limits_{1}^{x^2} \ln t dt\]

Answer 2

use integration by parts to find the integral of lnt with respect to t
then

put the limits :

int lnt dt = tlnt - t from 1 to x^2

ln1 - 1 - [ x^2 lnx^2 - x^2]

0 - 1 - x^2 lnx^2 + x^2


-1 - x^2 lnx^2 + x^2

Now differentiate this :


-2x lnx^2 - x^2 * 2x/x^2 + 2x

Answer 3

(-2 x log^2(x)-(x^2×2 x)/x^2+2 x) written liek this?

Answer 4

you can use Liebnitz - differentiation under the integral sign:

if:

\(\dfrac{dI}{dx} = \dfrac{d}{dx} \int\limits_{u(x)}^{v(x)} f(t) \; dt = \dfrac{d}{dx} \left[ F(v(x)) - F (u(x)) \right] \)

then \(\dfrac{dI}{dx} = f(v) \dfrac{dv}{dx} - f(u) \dfrac{du}{dx} \)

Answer 5

ok I got 2xln(x^2)

Answer 6

that's right!

Answer 7

Thnaks guys!