How do you integrate #int (3-2x)/(x(x²+3))# using partial fractions?

Question

Ezra Adams · Accepted Answer

#Lnx-1/2Ln(x^2+3)-(2sqrt3)/3*arctan(x/sqrt3)+C#

#int (3-2x)/[x*(x^2+3)*dx]#

I broke down the integrand into simpler fractions.

#(3-2x)/[x*(x^2+3)]#

=#A/x+(Bx+C)/(x^2+3)#

Following denominator expansion,

#A*(x^2+3)+(Bx+C)*x=3-2x#

#(A+B)*x^2+Cx+3A=3-2x#

After equating coefficients, #A+B=0#, #C=-2# and #3A=3#

From them, #A=1#, #B=-1# and #C=-2#

Thus,

#int (3-2x)/[x*(x^2+3)*dx]#

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

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

=#Lnx-1/2Ln(x^2+3)-(2sqrt3)/3*arctan(x/sqrt3)+C#

Luke Alvarez · Answer

undefined To integrate $\frac{3-2x}{x(x^2+3)}$ using partial fractions, first express the fraction as the sum of simpler fractions. Perform the decomposition as follows:

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

Solve for $A$, $B$, and $C$ by equating numerators:

$$3 - 2x = A(x^2 + 3) + (Bx + C)x$$

Equating coefficients of like terms:

$$3 - 2x = Ax^2 + 3A + Bx^2 + Cx$$

This equation holds for all $x$, so coefficients of like powers of $x$ must be equal:

For the $x^2$ terms:
$$A + B = 0$$

For the $x$ terms:
$$C = -2$$

For the constant terms:
$$3A = 3$$

Solve these equations to find $A$, $B$, and $C$:

$$A = 1$$
$$B = -1$$
$$C = -2$$

So, the partial fraction decomposition is:
$$\frac{3-2x}{x(x^2+3)} = \frac{1}{x} - \frac{x - 2}{x^2+3}$$

Now, integrate each term separately:

$$\int \frac{1}{x} dx - \int \frac{x - 2}{x^2+3} dx$$

$$\ln|x| - \frac{1}{2}\ln(x^2+3) - \sqrt{3}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$$

So, the integral of $\frac{3-2x}{x(x^2+3)}$ using partial fractions is:

$$\ln|x| - \frac{1}{2}\ln(x^2+3) - \sqrt{3}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$$