Question

how do i integrate this

Answer 1

x^2-4/x dx

Answer 2

divide first

Answer 3

i am assuming it is
\[\int\frac{x^2-4}{x}dx\]

Answer 4

can vi split them up into (x^2/x) - (4/x)?

Answer 5

then simplify into x - 4/x

Answer 6

divide and get
\[\int xdx -4\int \frac{1}{x}dx\]

Answer 7

yes, that is right , what you wrote

Answer 8

then it should be easy, right?

Answer 9

so no substitution?

Answer 10

oh no not at all

Answer 11

you can pretty much do it in your head
\[\frac{x^2}{2}-4\ln(x)\]

Answer 12

x^-1 is lnx right?

Answer 13

forgetting my rules

Answer 14

yes, the derivative of \(\ln(x)\) is \(\frac{1}{x}\) so the anti derivative of \(\frac{1}{x}\) is \(\ln(x)\)

Answer 15

what about x^2+2x+3/x^3+3x^2+9x

Answer 16

i solved this halfway

Answer 17

that might require a substitution

Answer 18

yeah, U=x^3+3x^2+9x and DU=3(x^2+2x+3)

Answer 19

put \(u=x^3+3x^2+9x\) gives ok you have it

Answer 20

i mutiply the inside my 3 and 1/3 on the outside

Answer 21

you only need to adjust
\[\frac{1}{3}du=(x^2+2x+3)dx\] and then you have
\[\frac{1}{3}\int \frac{du}{u}\]

Answer 22

yes, you have it

Answer 23

but then when i put it back in the 1/3\[\int\limits_{?}^{?}u^-1 \times du (3)\]

Answer 24

oops and the 1/3 on the outside

Answer 25

how would the 3 cancel out if the 1/3 is left in my answer?

Answer 26

yes, but most people just write
\[du=3(x^2+2x+3)dx\] so
\[\frac{1}{3}du=(x^2+2x+3)dx\] which is what you have

Answer 27

you are just adjusting by the constant. don't think too much about it
for example if you had
\[\int\sin(x^3)x^2dx\] you would say
\[u=x^3, du=3x^2dx, \frac{1}{3}du=x^2dx\] and then go right to
\[\frac{1}{3}\int\sin(u)du\]

Answer 28

as long as the powers work out, you can adjust by any constant

Answer 29

hmm okay

Answer 30

convince yourself that it is right by taking the derivative and seeing that it works out to what you want

Answer 31

alright thanks

Answer 32

yw