Question

Помогите решить интеграл

Answer 1

Ответ:

$\int\limits \frac{dx}{ {x}^{3} \sqrt{1 + {x}^{2} } } \\ \\ 1 + {x}^{2} = t \\ {x}^{2} = t - 1 \\2x dx = dt \\ dx = \frac{dt}{2x} \\ \\ \int\limits \frac{dt}{ {x}^{3} \times 2 {x}^{} \sqrt{t} } = \frac{1}{2} \int\limits \frac{dt}{ \sqrt{t} \times {(t - 1)}^{2} } \\ \\ \sqrt{t} = z \\ \frac{1}{2 \sqrt{t} } dt = dz \\ dt = 2 \sqrt{t} dz = 2zdz \\ \\ \frac{1}{2} \int\limits \frac{2zdz}{z( {z}^{2} - 1) {}^{2} } = \frac{2}{2} \int\limits \frac{dz}{ {(z {}^{2} - 1) }^{2} } = \\ = \int\limits \frac{dz}{ {(z - 1) {}^{2} (z + 1)}^{2} }$

С помощью неопределенных коэффициентов:

$\frac{1}{ {(z - 1)}^{2} {( z+ 1)}^{2} } = \frac{a}{ z+ 1} + \frac{b}{(z + 1) {}^{2} } + \frac{c}{z - 1} + \frac{d}{ {( z- 1)}^{2} } \\ 1 = a(z + 1)(z - 1) {}^{2} + b {(z - 1)}^{2} + c(z - 1)(z + 1) {}^{2} + d(z + 1) {}^{2} \\ 1 = a(z {}^{2} - 1)(z - 1) + b(z {}^{2} - 2 z + 1) + c(z {}^{2} - 1)( z + 1) + d( {z}^{2} + 2z + 1) \\ 1 = az {}^{3} - az {}^{2} - az + a + b {}^{} z {}^{2} - 2b z + b + cz {}^{3} + cz {}^{2} - cz - c + dz {}^{2} + 2 dz + d \\ \\ 0 = a + c \\ 0 = - a + b + c + d \\ 0 = - a - 2b - c + 2d\\ 1 = a + b - c + d \\ \\ a = b = d = \frac{1}{4} \\ c= - \frac{1}{4}$

$\frac{1}{4} (\int\limits \frac{dz}{z + 1} + \int\limits \frac{dz}{(z + 1) {}^{2} } - \int\limits \frac{dz}{z - 1} + \int\limits \frac{dz}{(z - 1) {}^{2} } ) = \\ = \frac{1}{4} ( ln( |z + 1| ) - \frac{1}{z + 1} - ln( |z - 1| ) - \frac{1}{z - 1} ) + C = \\ = \frac{1}{4} ln( | \frac{z + 1}{z - 1} | ) - \frac{1}{4(z + 1)} - \frac{1}{4(z - 1)} + C = \\ = \frac{1}{4} ln( | \frac{ \sqrt{t} + 1}{ \sqrt{t} - 1} | ) - \frac{1}{4( \sqrt{t} + 1) } - \frac{1}{4( \sqrt{t} - 1)} + C= \\ = \frac{1}{4} ln( | \frac{ \sqrt{1 + {x}^{2} } + 1}{ \sqrt{1 + {x}^{2} } - 1 } | ) - \frac{1}{4( \sqrt{1 + {x}^{2} } + 1) } - \frac{1}{4( \sqrt{1 + {x}^{2} } - 1) } + C$