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\(\int \frac{x^4+1}{x^3-x^2+x-1}dx\)
całka
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\(\frac{x^4+1}{x^3-x^2+x-1}=\frac{x(x^3-x^2+x-1)+x^2-x^2+x+1}{x^2-x^2+x-1}=x+\frac{x^2-x^2+x-1+2}{x^3-x^2+x-1}=x+1-\frac{2}{(x-1)(x^2+1)}\)
\(\frac{2}{(x^2+1)(x-1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}=\frac{(A+C)x^2+(B-A)x+(C-B)}{(x^2+1)(x-1)}=\\ \left(\begin{cases}A+C=0\\B-A=0\\C-B=2 \end{cases} \\ \begin{cases}A=-1\\B=-1\\C=1 \end{cases} \right) \\=\frac{1}{x-1}-\frac{x+1}{x^2+1}=\frac{1}{x-1}-\frac{x}{x^2+1}-\frac{1}{x^2+1}\)
\(\frac{x^4+1}{x^3-x^2+x-1}=x+1+\frac{1}{x-1}-\frac{x}{x^2+1}-\frac{1}{x^2+1}\)
\(\int\ \frac{x^4+1}{x^3-x^2+x-1}dx=\int\ x\ dx+\int\ dx+\int\ \frac{1}{x-1}dx-\int\ \frac{1}{1+x^2}dx-\frac{1}{2}\int\frac{2x}{x^2+1}dx=\\=\frac{1}{2}x^2+x+ln|x-1|-\frac{1}{2}ln|x^2+1|-arc\ tgx+C=\\=\frac{1}{2}x^2+x+ln\frac{|x-1|}{\sqrt{x^2+1}}-arc\ tg\ x+C\)
\(\frac{2}{(x^2+1)(x-1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}=\frac{(A+C)x^2+(B-A)x+(C-B)}{(x^2+1)(x-1)}=\\ \left(\begin{cases}A+C=0\\B-A=0\\C-B=2 \end{cases} \\ \begin{cases}A=-1\\B=-1\\C=1 \end{cases} \right) \\=\frac{1}{x-1}-\frac{x+1}{x^2+1}=\frac{1}{x-1}-\frac{x}{x^2+1}-\frac{1}{x^2+1}\)
\(\frac{x^4+1}{x^3-x^2+x-1}=x+1+\frac{1}{x-1}-\frac{x}{x^2+1}-\frac{1}{x^2+1}\)
\(\int\ \frac{x^4+1}{x^3-x^2+x-1}dx=\int\ x\ dx+\int\ dx+\int\ \frac{1}{x-1}dx-\int\ \frac{1}{1+x^2}dx-\frac{1}{2}\int\frac{2x}{x^2+1}dx=\\=\frac{1}{2}x^2+x+ln|x-1|-\frac{1}{2}ln|x^2+1|-arc\ tgx+C=\\=\frac{1}{2}x^2+x+ln\frac{|x-1|}{\sqrt{x^2+1}}-arc\ tg\ x+C\)