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[haharuka]

\(\int_{}^{} \frac{x}{2x^3-x^2-4x+3}\)

[eresh]

\(\int_{}^{} \frac{x}{2x^3-x^2-4x+3}\)

\(\int_{}^{} \frac{x}{2x^3-x^2-4x+3}dx=\int\frac{x}{2(x+\frac{3}{2})(x-1)^2}dx\)

\(\frac{x}{2(x+\frac{3}{2})(x-1)^2}=\frac{A}{2x+3}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\\
x=A(x-1)^2+B(x-1)(2x+3)+C(2x+3)\\
x=Ax^2-2Ax+A+2Bx^2+Bx-3B+2Cx+3C\\
\begin{cases}A+2B=0\\-2A+B+2C=1\\A+3C=0\end{cases}\\
\begin{cases}A=-\frac{6}{25}\\B=\frac{3}{25}\\C=\frac{1}{5}\end{cases}\)
\(I=\int\frac{-\frac{6}{25}dx}{2x-3}+\int\frac{\frac{3}{25}dx}{x-1}+\int\frac{\frac{1}{5}dx}{(x-1)^2}\\
I=-\frac{3}{25}\ln|x-\frac{3}{2}|+\frac{3}{25}\ln|x-1|-\frac{1}{5}\cdot\frac{1}{x-1}+C\)

[haharuka]

Jest jakiś sprawdzony sposób jak wpaść na takie rozwiązanie?