How do you integrate #(3x) / (x^2 * (x^2+1) )# using partial fractions?

Question

Ezra Adams · Accepted Answer

#3/2ln{x^2/(x^2+1)}+C.#

Let #I=int(3x)/{x^2(x^2+1)}dx#

Substitute #x^2=t," so that, "2xdx=dt, or, xdx=1/2dt#

#:. I=3int(1/2)/{t(t+1)}dt=3/2int{(t+1)-t}/{t(t+1)}dt#

#=3/2int[(t+1)/{t(t+1)}-t/{t(t+1)}]dt#

#=3/2int[1/t-1/(t+1)]dt#

#=3/2[ln|t|-ln|t+1|]#

#=3/2ln|t/(t+1)|, and, because, t=x^2,#

#I=3/2ln{x^2/(x^2+1)}+C.#

Have fun with math!

Ezekiel Alexander · Answer

undefined To integrate $\frac{3x}{x^2(x^2+1)}$ using partial fractions, we first express the fraction as a sum of simpler fractions:

$\frac{3x}{x^2(x^2+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+1}$

We then find the values of $A$, $B$, $C$, and $D$ by equating numerators:

$3x = A(x^2)(x^2+1) + B(x)(x^2+1) + (Cx+D)(x^2)$

Next, we clear the denominators and group like terms:

$3x = Ax^4 + Ax^2 + Bx^3 + Bx + Cx^3 + Dx^2$

Comparing coefficients on both sides, we get the following system of equations:

$A = 0$  
$B + C = 0$  
$A + D = 3$  
$B = 0$

Solving the system, we find $A = 0$, $B = 0$, $C = 0$, and $D = 3$.

Therefore, the partial fraction decomposition is:

$\frac{3x}{x^2(x^2+1)} = \frac{3}{x^2+1}$

Now, we can integrate each term separately:

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

Using the substitution $u = x^2 + 1$, we get $du = 2x \, dx$, which gives us:

$\frac{3}{2} \int \frac{1}{u} \, du = \frac{3}{2} \ln|u| + C = \frac{3}{2} \ln|x^2+1| + C$

Therefore, the integral of $\frac{3x}{x^2(x^2+1)}$ is $\frac{3}{2} \ln|x^2+1| + C$.