Question

Calculus - Parts - Integration

Answer 1

\[\int\limits_{1}^{e} \ln(x^2)dx\]

Answer 2

let u = ln(x^2) and dv = dx


did you try that??

Answer 3

\[du = \frac{1}{x^2} (2xdx)\] then \[v = x\]

Answer 4

do you need more help??? @emily.polhemus

Answer 5

Yes please, I just need help finishing it

Answer 6

well what do you think is the next step here?

Answer 7

wait wouldn't du = (1/x^2)dx not du=1/x^2(2xdx) because that would just be (2/x)dx?

Answer 8

remember chain rule? derivative of ln u = 1/u (du)

du means derivative of u


in this case u = x^2 so ln(x^2) is equal to 1/x^2 then multiplied to the derivative of x^2 which is 2x

therefore we have \[\frac{2xdx}{x^2}\] got it?

Answer 9

so it reduces to \[(2/x)dx\] ?

Answer 10

yep

Answer 11

so du = 2dx/x

Answer 12

Ohhh I think I got it now, thanks!

Answer 13

Wait, hold on

Answer 14

sure

Answer 15

Thank you so much!

Answer 16

If I have a similar question could you please help me?

Answer 17

im going there..

Answer 18

If you're finding it hard choosing you're u, LIATE/LIAET should help you.
L-logarithmic functions
I-Inverse functions
A-Algebraic functions
E-exponential functions
T-Trigonometric functions
Preference to your choice of u increases as you move down. So if you had \[\int\limits_{}^{} t^2 e^t dt\]
You would choose you're u as \[t^2\] since algebraic functions come before exponential functions

Answer 19

your* u