Question

calculate $int\frac{1x^{2}}{(x - 3)(x + 2)^{2}}dx$. page generated september 12, 2025, 12:17:13 pm edt ork © 1996 - 2025 | theme: math4 | ww_version: 2.20 | pg_version 2.20 the webwork project

Explanation:

Step1: Perform partial - fraction decomposition

Let $\frac{x^{2}}{(x - 3)(x + 2)^{2}}=\frac{A}{x - 3}+\frac{B}{x+2}+\frac{C}{(x + 2)^{2}}$. Then $x^{2}=A(x + 2)^{2}+B(x - 3)(x + 2)+C(x - 3)$.
If $x = 3$, we have $3^{2}=A(3 + 2)^{2}$, so $9 = 25A$, and $A=\frac{9}{25}$.
If $x=-2$, we have $(-2)^{2}=C(-2 - 3)$, so $4=-5C$, and $C =-\frac{4}{5}$.
Expand the right - hand side: $x^{2}=A(x^{2}+4x + 4)+B(x^{2}-x - 6)+C(x - 3)$.
$x^{2}=(A + B)x^{2}+(4A - B + C)x+(4A-6B - 3C)$.
Since $A=\frac{9}{25}$ and $C =-\frac{4}{5}$, and comparing the coefficients of $x^{2}$: $1=A + B$, so $B=1 - A=1-\frac{9}{25}=\frac{16}{25}$.
So $\frac{x^{2}}{(x - 3)(x + 2)^{2}}=\frac{9}{25(x - 3)}+\frac{16}{25(x + 2)}-\frac{4}{5(x + 2)^{2}}$.

Step2: Integrate term - by - term

$\int\frac{x^{2}}{(x - 3)(x + 2)^{2}}dx=\int\frac{9}{25(x - 3)}dx+\int\frac{16}{25(x + 2)}dx-\int\frac{4}{5(x + 2)^{2}}dx$.
$\int\frac{9}{25(x - 3)}dx=\frac{9}{25}\ln|x - 3|+C_1$.
$\int\frac{16}{25(x + 2)}dx=\frac{16}{25}\ln|x + 2|+C_2$.
For $\int\frac{4}{5(x + 2)^{2}}dx$, let $u=x + 2$, $du=dx$, then $\int\frac{4}{5u^{2}}du=-\frac{4}{5u}+C_3=-\frac{4}{5(x + 2)}+C_3$.

Step3: Combine the results

$\int\frac{x^{2}}{(x - 3)(x + 2)^{2}}dx=\frac{9}{25}\ln|x - 3|+\frac{16}{25}\ln|x + 2|+\frac{4}{5(x + 2)}+C$, where $C = C_1+C_2+C_3$.

Answer:

$\frac{9}{25}\ln|x - 3|+\frac{16}{25}\ln|x + 2|+\frac{4}{5(x + 2)}+C$