How do you integrate #int(2x +3)/(x^4-9x^2)# using partial fractions?

Question

Emily Ammerman · Accepted Answer

#I=1/18ln|x+3|+1/6ln|x-3|-2/9ln|x|+1/(3x)+c#

Here,

#I=int(2x+3)/(x^4-9x^2)dx=int(2x+3)/(x^2(x^2-9))dx#

#I=int(2x+3)/(x^2(x-3)(x+3))dx#

Let,

#(2x+3)/(x^2(x-3)(x+3))=A/x+B/x^2+C/(x-3)+D/(x+3)#

#2x+3#=#Ax(x^2-9)+B(x^2-9)+Cx^2(x+3)+Dx^2(x-3)#

#2x+3#=#x^3(A+C+D)+x^2(B+3C-3D)+x(-9A)-9B#

#"Comparing co efficients of "# #x^3,x^2,x and# #"constant term :"#

#A+C+D=0..........to(1)#

#B+3C-3D=0.......to(2)#

#-9A=2=>color(red)(A=-2/9to(3)#

#-9B=3=>color(red)(B=-1/3to(4)#

From #(1) and (3)color(white)(..........)#From #(2) and(4)#

#C+D=2/9to(5)color(white)(.......)3C-3D=1/3=>C-D=1/9to(6)#

Adding #(5) and (6)# we get

#C+C=2/9+1/9=>2C=3/9=>C=3/18=>color(red)(C=1/6#

From #(5)# ,we get

#1/6+D=2/9=>D=2/9-1/6=(4-3)/18=>color(red)(D=1/18#

So,

#I=int[(-2/9)/x+(-1/3)/x^2+(1/6)/(x-3)+(1/18)/(x+3)]dx#

#=-2/9ln|x|-1/3(x^(-2+1)/(-2+1))+1/6ln|x-3|+1/18ln|x+3|# + c

#=-2/9ln|x|-1/3(x^(-1)/(-1))+1/6ln|x-3|+1/18ln|x+3|+c#

#=1/18ln|x+3|+1/6ln|x-3|-2/9ln|x|+1/(3x)+c#

Paisley Johnson · Answer

undefined To integrate $\frac{{2x + 3}}{{x^4 - 9x^2}}$ using partial fractions, you first factor the denominator, $x^4 - 9x^2$, as $x^2(x^2 - 9)$. Then, you express the fraction as $\frac{{2x + 3}}{{x^2(x + 3)(x - 3)}}$.

The partial fraction decomposition of this expression is: 
$$\frac{{2x + 3}}{{x^2(x + 3)(x - 3)}} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 3} + \frac{D}{x - 3}$$

Multiplying both sides by the common denominator $x^2(x + 3)(x - 3)$, you get:
$$2x + 3 = A(x + 3)(x - 3) + B(x - 3) + Cx(x - 3) + Dx(x + 3)$$

By equating coefficients of like terms, solve for $A$, $B$, $C$, and $D$. Once you find these constants, you integrate each term separately, and then sum up the integrals.

The integral of $\frac{A}{x}$ is $A\ln|x|$, the integral of $\frac{B}{x^2}$ is $-\frac{B}{x}$, the integral of $\frac{C}{x + 3}$ is $C\ln|x + 3|$, and the integral of $\frac{D}{x - 3}$ is $D\ln|x - 3|$.

After integrating each term, combine them to get the final result.