Question

intergrate (x)cube/(x)sq - 1

Answer 1

\[\int\limits_{}^{} x^{3} / (x^{2} - 1)\]

Answer 2

\[\int \frac{x^3}{x^2 - 1}\]
divide this thing first

x - x/x^2 - 1
________
x^2 - 1 | x^3
x^3
-------
0 - x
now sub

\[\int (x - \frac{x}{x^2 - 1} )dx\]
distribute the integral
\[\int xdx - \int \frac{x}{x^2 - 1}dx\]
\[\int xdx - \frac 12 \int \frac{1}{u - 1}du\]

Answer 3

can anyone explain the 2nd part to the solution provided by lgbastalloe?

Answer 4

division of polynomials?

Answer 5

or the substitution?

Answer 6

the substitution

Answer 7

you got the division though right?

Answer 8

yep i got it. only the substituition.

Answer 9

if i had \[\int \frac{4}{2}dx\]then i can just write this as \[\int 2dx\] right? since the quotient of 4 divided by 2 is 2

same here the quotient of \[\frac{x^3}{x^2 - 1} \implies x - \frac{x}{x^2 - 1}\]
make sense?

Answer 10

yes. i get that. but how did you substitute U into the equation?

Answer 11

ohh u...

Answer 12

when i did it, i got stuck. i couldnt get the same outcome as you.

Answer 13

\[\int \frac{x}{x^2 - 1}dx\]
let u = x^2 - 1
du = 2x dx
du/2 = xdx

\[\int \frac{x}{x^2 - 1}dx \implies \frac{1}{2} \int \frac{du}{u}\]
lol just noticed i made a mistake in what i write a while ago

Answer 14

OH Thanks alot. i get it now.

Answer 15

\[\huge \color{darkblue}{\textbf{<tips hat>}}\]