Question

Помогите интегрировать дробь

Answer 1

Ответ:

$\int\limits \frac{2 {x}^{3} + 6 {x}^{2} + 7x + 2}{x {(x + 1)}^{3} } dx$

С помощью неопределенных коэффициентов разделим на простейшие дроби:

$\frac{2 {x}^{3} + 6 {x}^{2} + 7x + 2 }{x {(x + 1)}^{3} } = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{ {(x + 1)}^{2} } + \frac{D}{ {(x + 1)}^{3} } \\ 2 {x}^{3} + 6 {x}^{2} + 7x + 2 = A {(x + 1)}^{3} +Bx {(x + 1)}^{2} +Cx(x + 1) + Dx \\ 2 {x}^{3} + 6 {x}^{2} + 7x + 2 = A( {x}^{3} + 3x + 3 {x}^{2} + 1) + Bx( {x}^{2} + 2x + 1) + C {x}^{2} + Cx + Dx \\ 2 {x}^{3} + 6 {x}^{2} + 7x + 2 = A{x}^{3} + 3A {x}^{2} + 3Ax + A + B{x}^{3} + 2B {x}^{2} + Bx + C {x}^{2} + Cx + Dx \\ \\ \begin{cases}2 = A + B& \\6 = 3A + 2B+ C & \\ 7 = 3A + B + C + D \\ 2 = A\end{cases} \\ \\ \begin{cases}A = 2 & \\B = 2 - 2 = - 0& \\ C= 6 - 6 - 0 = 0\\ D = 7 - 6 - 1 = 1\end{cases} \\ \\ \int\limits \frac{2dx}{x} + \int\limits \frac{dx}{ {(x + 1)}^{3} } = 2ln |x| + \int\limits \frac{d(x + 1)}{ {(x + 1)}^{3} } = \\ = 2ln |x| + \frac{ {(x + 1)}^{ - 2} }{( - 2)} + C = 2ln |x| - \frac{1}{2 {(x + 1)}^{2} } + C$