En aquest cas la factorització del polinomi denominador \(B(x)\) serà de la forma:
\(\displaystyle B(x)= \left( x-a_1 \right)^{m_1}\cdot\left( x-a_2 \right)^{m_2}\ldots\cdot\left( x-a_p \right)^{m_p} \)
on:
\(m_1+m_2+\ldots+m_p=n\)
La fracció algebraica \(\displaystyle \frac{A(x)}{B(x)}\) es pot descompondre de manera única com a combinació lineal de fraccions simples. En aquesta descomposició cada arrel \(a_i\) dóna lloc a una suma de \(m_i\) fraccions algebraiques de la forma:
\(\displaystyle \frac{M_{i1}}{x-a_i}+\frac{M_{i2}}{\left(x-a_i\right)^2}+\ldots+\frac{M_{im_i}}{\left(x-a_i\right)^{m_i}}\)
Per tant:
\(\displaystyle \begin{align} \frac{A(x)}{B(x)}= &\frac{M_{11}}{x-a_1}+\frac{M_{12}}{\left(x-a_1\right)^2}+\ldots+\frac{M_{1m_1}}{\left(x-a_1\right)^{m_1}}\\ &+\frac{M_{21}}{x-a_2}+\frac{M_{22}}{\left(x-a_2\right)^2}+\ldots+\frac{M_{2m_1}}{\left(x-a_2\right)^{m_2}}\\ &+\ldots\\ &+\frac{M_{q1}}{x-a_q}+\frac{M_{q2}}{\left(x-a_q\right)^2}+\ldots+\frac{M_{qm_1}}{\left(x-a_q\right)^{m_q}} \end{align} \)
Totes aquestes fraccions algebraiques són integrables immediatament.
Exemple
Volem calcular la integral:
\(\displaystyle \int \frac{9}{x^3-3x^2} \,\mathrm{d}x\)
Primer factoritzem el denominador:
\(\displaystyle \begin{align} x^3-3x^2=x^2(x-3) \quad &\Rightarrow\quad \frac{9}{x^2(x-3)}=\frac{A}{x-3}+\frac{B}{x}+\frac{C}{x^2}\\[8pt] &\Rightarrow\quad \frac{9}{x^2(x-3)}=\frac{Ax^2+Bx(x-3)+C(x-3)}{x^2(x-3)}\\[8pt] &\Rightarrow\quad \frac{9}{x^2(x-3)}=\frac{(A+B)x^2+(-3B+C)x-3C}{x^2(x-3)}\\[8pt] &\Rightarrow\quad 9=(A+B)x^2+(-3B+C)x-3C\\[8pt] &\Rightarrow\quad \left\lbrace\begin{array}{r} A+B=0 \\ -3B+C=0 \\ -3C=9 \end{array} \right. \\[8pt] &\Rightarrow\quad \left\lbrace\begin{array}{l} A=1 \\ B=-1 \\ C=-3 \end{array} \right. \\[8pt] \end{align} \)
I ja podem resoldre la integral:
\(\displaystyle \begin{align} \int \frac{9}{x^3-3x^2} \,\mathrm{d}x &= \int \frac{1}{x-3} \,\mathrm{d}x - \int \frac{1}{x} \,\mathrm{d}x - 3 \int \frac{1}{x^2} \,\mathrm{d}x\\[6pt] &= \ln \left\lvert x-3 \right\rvert - \ln \left\lvert x \right\rvert + \frac{3}{x} + C \end{align} \)
Exercici 22
Calcula la integral indefinida \(\displaystyle \int \frac{4x+4}{x^3-4x^2+4x}\,\mathrm{d}x \).
Solució:Exercici 23
Calcula la integral indefinida \(\displaystyle \int \frac{2x+3}{x^4+x^3}\,\mathrm{d}x \).
Solució: