Question

integral of (x^2-x+6)/(x^3 + 3x)

Answer 1

by dividing then using partial fractions please

Answer 2

you cannot divide so just use partial fractions

Answer 3

can you go through the steps for me please?

Answer 4

\[\frac{x^2-x+6}{x(x^2+3)}\]

Answer 5

sorry i am just figuring it out on paper, the reason you cannot long divide is because the degree is larger on the bottom then the top

Answer 6

\[\frac{x^2-x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}\]

Answer 7

okay i've gotten that far

Answer 8

wow my computer is lagggy. okay \[x^2-x+6=A(x^2+3)+Bx^2+Cx\]

Answer 9

set x=0 and solve 6=3A therfore A=2, now you can sub 2 in for A and move it over to the left hand side \[x^2-x+6=2x^2+6+Bx^2+Cx..........-x^2-x=Bx^2+Cx..\] from the equation you can see that the only way the sides could be equal is if B=-1 and C=-1

Answer 10

does it make sense so far? or is something not working?

Answer 11

\[\int\limits \frac{2}{x}dx+\int\limits \frac{-x}{x^2+3}dx- \int\limits \frac{1}{x^2+3} dx\]

Answer 12

the first integral is 2ln(x).........the second integral let u=x^2+3........... -1/2du=-xdx..........and it ends up being -1/2ln(x^2+3).........the third integral recall tan^-1 = 1/(x^2+1)

Answer 13

\[\tan^-1 \frac{\frac{x}{\sqrt{3}}}{\sqrt{3}}\]