Integral (x^2-x+8)/(x^3+2x) dx
I believe we have to do partial fraction decomposition to start this, but I'm not sure which partial fraction decomposition rule this would fall under.
Any help in the right direction would be appreciated!
Integral (x^2-x+8)/(x^3+2x) dx
I believe we have to do partial fraction decomposition to start this, but I'm not sure which partial fraction decomposition rule this would fall under.
Any help in the right direction would be appreciated!
Since the degree of the numerator is less than that of the denominator, you do not need to do a polynomial division, so you can proceed to the next step, which is to factor the denominator and set up the partial fraction decomposition:
$\displaystyle x^3+2x = x(x^2+2)$, and $\displaystyle x^2+2$ can not be factored, so write:
$\displaystyle \frac{x^2-x+8}{x^3+2x}=\frac{A}{x}+\frac{Bx+C}{x^2+2}$
- Hollywood
Thanks! I have to find out what A,B,C equals. So I'll do when x=0,1,2 So A=4 when x=0. I dont understand this part: why does 1=A+B and B=-1 and therefore C=-1?
I just went on to solve and got the integral of 4/x + -3x-1/x^2+2
which simplifies to to integral (4/x -3 x/x^2+2 -1/x^2+2) dx
that equals 4ln|x|-(3/2)ln(x^2+2)-(1/sqrt2)tan^-1(x/sqrt2)
the bolded parts I would really appreciate any sort of simple explanation (like teaching it to a 5 yr old hahaha). thanks!
What I would do from $\displaystyle \displaystyle x^2 - x + 8 = A\left( x^2 + 2 \right) + x \left( B\,x + C \right) $ would be to expand all the brackets on the right, simplify, then equate like powers of x.
$\displaystyle \displaystyle \begin{align*} x^2 - x + 8 &= A\, x^2 + 2A + B\,x^2 + C\,x \\ 1x^2 + \left( -1 \right) x + 8 &= \left( A + B \right) x^2 + C\, x + 2A \\ A + B = 1 \textrm{ and } C = -1 \textrm{ and } 2A &= 8 \\ A = 4 \textrm{ and } B = -3 \textrm{ and } C &= -1 \end{align*} $
So that means
$\displaystyle \displaystyle \begin{align*} \int{\frac{x^2 - x + 8}{x^3 + 2x}\,dx} &= \int{ \frac{4}{x} + \frac{-3x - 1}{x^2 + 2} \,dx } \\ &= \int{\frac{4}{x}\,dx} - \int{\frac{3x}{x^2 + 2} \,dx} - \int{ \frac{1}{x^2 + 2} \,dx } \\ &= \int{\frac{4}{x}\,dx} - \frac{3}{2}\int{\frac{2x}{x^2 + 2}\,dx} - \int{ \frac{1}{x^2 + 2}\,dx } \end{align*}$
Now the first integral is a logarithm, the second is solved using a substitution $\displaystyle \displaystyle u = x^2 + 2 \implies du = 2x\,dx$ and the third is solved using a substitution $\displaystyle \displaystyle x = \sqrt{2}\tan{(\theta)} \implies dx = \sqrt{2}\,sec^2{(\theta)}\,d\theta$.