Question

i need help with integrals, i will attach the 2 problems, the first one has to do with substitution and the second one with "by parts".

Answer 1

here they are :)

Answer 2

Okay, I can help you with all your integral fears :P

Answer 3

thank u!

Answer 4

Okay, the first one you want to use ln(x)=u
Differentiating that you have:
(1/x)dx=du
You notice that you have an x in the denominator.
So replace the integral with:
\[7\int\limits \frac{du}{u^3}=7 \int\limits u^{-3}du=\frac{-7}{2}u^{-2}+C\]
But u=ln(x) so you have:
\[\frac{-7}{2\ln^2(x)}+C\]

Answer 5

Does that make sense before I go to the second one?

Answer 6

i think, its just that my teacher uses z and dz, instead of u and du. thank u:)

Answer 7

Just a change of variables, you could use anything you wanted :P

Answer 8

ok.

Answer 9

Ready for the next one?

Answer 10

yes

Answer 11

Okay, for this one you want to pick a u that is easy to differentiate, and a dv that is easy to integrate.
So for this problem:
u=ln(x) dv=1/x^3dx
du=1/x dx v=-1/2x^2
Using the formula:
\[uv-\int\limits v du\]
You have:
\[\frac{-\ln(x)}{2x^2}+\int\limits \frac{dx}{x^3}=\frac{-\ln(x)}{2x^2}-\frac{1}{4x^2}+C\]

Answer 12

There is a "simpler" integration by parts approach but its slightly more "difficult".

Answer 13

could i have done this instead...
ln x(-x^-2/2)-∫-x^-2/2 (1/x) dx = ln x(-x^-2/2)-x^-2/4+C

Answer 14

That's exactly right :) Its more important to understand how to approach the problems than actually working them. Good job :D

Answer 15

thank u, so it would be fine if instead i put the negative sign inside the parentheses, like this:
ln x (-x^-2/2)... instead of: -ln x (x^-2/2)...

Answer 16

It doesn't matter. They are equivalent. Most people put it outside the ln but there is not real difference.

Answer 17

ok thank you very much. i had a question about the first one...is it the same too, if i put the exponent in different places: -7/2(lnx)^2+C instead of: -7/2ln^2x +C ??

Answer 18

There is no difference either :P

Answer 19

thanks!