интеграл............................... Created by uh19. algebra-ru

Решение смотри на фотографии

интеграл...

manyny06

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$\int {\frac{2x^2 -1}{x^3+6x^2 +9x}} \, dx = \int {\frac{2x^2 -1}{x \cdot (x^2+6x +9)}} \, dx = \int {\frac{2x^2 -1}{x \cdot (x+3)^2}} \, dx = (*) \\ \\ \frac{A}{x} + \frac{B}{x+3}+ \frac{C}{(x+3)^2}=\frac{2x^2-1}{x \cdot (x+3)^2} \\ \\ A(x^2 +6x+9) + B(x^2+3x) +Cx=2x^2-1$

$\left\{\!\begin{aligned}
& A +B=2 \\
& 6A+3B+C=0 \\
& 9A =-1
\end{aligned}\right. \ \ \left\{\!\begin{aligned}
& B = 2+\frac{1}{9}=\frac{19}{9} \\
& C=\frac{6}{9}-\frac{3 \cdot 19}{9}=\frac{2-19}{3}=-\frac{17}{3} \\
& A =-\frac{1}{9}\end{aligned}\right. \\ \\ \\ \int {(-\frac{1}{9x} + \frac{19}{9 \cdot (x+3)} - \frac{17}{3 \cdot (x+3)^2}}}) \, dx= -\frac{1}{9} \int \frac{dx}{x} + \frac{19}{9} \int \frac{dx}{x+3} - \frac{17}{3} \int \frac{dx}{x+3)^2} =$

$\\\\ =-\frac{1}{9} \ln{|x|} + \frac{19}{9} \ln{|x+3|} - \frac{17}{3} \cdot (- \frac{1}{x+3}) + C = \\ \\ = -\frac{1}{9} \ln{|x|} + \frac{19}{9} \ln{|x+3|} + \frac{17}{ 3 \cdot (x+3)}+ C$

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Verified answer

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1 2 3 64b680cd307ab

bd1cfa560ea8d3bf4201d287e488675d

vi ii 64afbe65f0afb

1 2 3 4 64abe4bbc9156

1 2 3 4 64a9f1c9b84fa

ed7553109b051819d37f06608bb610ab

1 2 ii 64a626991420a

tex]

1 a xy b b xy a 2b ta b 64a352121ddb9

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