Question

математический анализ

Answer 1

Ответ:

1)

$\boldsymbol{\boxed{\int {\frac{\cos x}{\sin^{5} x} } \, dx =-\frac{1}{4 \sin^{4} x} + C}}$

2)

$\boldsymbol{\boxed{ \int {\frac{x+2}{x^{2} +4x+5} } \, dx =\dfrac{1}{2} \ln |x^{2} +4x+5 |+C}}$

3)

$\boldsymbol{\boxed{\int {\frac{e^{\sqrt{x} }}{\sqrt{x} } } \, dx = 2e^{\sqrt{x}} +C}}$

4)

$\boldsymbol{\boxed{ \int\limits^{0}_{-1} {(2x+3)e^{-x}} \, dx =3e -5 }}$

5)

$\boldsymbol{\boxed{\int {xe^{2x}} \, dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} +C}}$

Примечание:

По таблице интегралов:

$\boxed{\displaystyle \int x^{n} \ dx = \frac{x^{n + 1}}{n + 1} + C; n \neq -1, x > 0}$

$\boxed{\int \frac{dx}{x} = \ln|x|+C }$

$\boxed{\int {e^{x}} \, dx = e^{x}+C }$

По свойствам интегралов:

$\boxed{ \displaystyle \int \sum\limits_{i=1}^n {C_{i}f_{i}(x)} \, dx = \sum\limits_{i=1}^nC_{i} \int {f_{i}(x)} \, dx}$

Интегрирование по частям:

$\boxed{\int {u} \, dv = uv - \int {v} \, du}$

$u,v \ -$ функции от аргумента x

Пошаговое объяснение:

1)

$\displaystyle \int {\frac{\cos x}{\sin^{5} x} } \, dx = \int {\sin^{-5} x} \, d(\sin x) = \frac{\sin^{-5+1} x}{-5+1} +C = -\frac{1}{4 \sin^{4} x} + C$

2)

$\displaystyle \int {\frac{x+2}{x^{2} +4x+5} } \, dx = \frac{1}{2} \int {\frac{2x+4}{x^{2} +4x+5} } \, dx = \frac{1}{2} \int {\frac{d(x^{2} +4x+5)}{(x^{2} +4x+5)} }=$

$=\dfrac{1}{2} \ln |x^{2} +4x+5 |+C$

3)

$\displaystyle \int {\frac{e^{\sqrt{x} }}{\sqrt{x} } } \, dx = 2\int {e^{\sqrt{x}} \, d(\sqrt{x} ) = 2e^{\sqrt{x}} +C$

4)

$\displaystyle \int\limits^{0}_{-1} {(2x+3)e^{-x}} \, dx = \int\limits^{0}_{-1} {(2xe^{-x}+3e^{-x})} \, dx = \int\limits^{0}_{-1} {2xe^{-x}} \, dx +\int\limits^{0}_{-1} {3e^{-x}} \, dx =$

$\displaystyle = 2\int\limits^{0}_{-1} {xe^{-x}} \, dx +3\int\limits^{0}_{-1} {e^{-x}} \, dx =$

а)

$\displaystyle\int {xe^{-x}} \, dx=$

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$u = x \Longrightarrow du = dx$

$\displaystyle dv = e^{-x} \ dx \Longrightarrow v = \int {e^{-x}} \, dx =-\int {e^{-x}} \, d(-x) =-e^{-x}$

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$\displaystyle =-xe^{-x} - \int {-e^{-x}} \, dx = -xe^{-x} + \int {e^{-x}} \, dx =-xe^{-x} -e^{-x}+C$

$\displaystyle\int\limits^{0}_{-1} {xe^{-x}} \, dx= \bigg(-xe^{-x} -e^{-x} \bigg) \bigg |_{-1}^{0} = \bigg(-e^{-x}(x+1) \bigg) \bigg |_{-1}^{0} = \bigg(-\frac{x+1}{e^{x}} \bigg) \bigg |_{-1}^{0} =$

$\displaystyle = \bigg(-\frac{0+1}{e^{0}} \bigg) - \bigg(-\frac{-1+1}{e^{-1}} \bigg) = -1 - 0 =-1$

б)

$\displaystyle\int\limits^{0}_{-1} {e^{-x}} \, dx = -e^{-x} \bigg |_{-1}^{0} = -\frac{1}{e^{x}} \bigg |_{-1}^{0} = -\frac{1}{e^{0}} + -\frac{1}{e^{-1}} = -\frac{1}{1} + e =-1+e = e -1$

в)

$\displaystyle \int\limits^{0}_{-1} {(2x+3)e^{-x}} \, dx = 2\int\limits^{0}_{-1} {xe^{-x}} \, dx +3\int\limits^{0}_{-1} {e^{-x}} \, dx = 2 \cdot (-1) + 3(e -1) =$

$= -2 + 3e -3 = 3e -5$

5)

$\displaystyle \int {xe^{2x}} \, dx =$

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$u = x \Longrightarrow du = dx$

$\displaystyle dv = e^{2x} \ dx \Longrightarrow v = \int {e^{2x}} \, dx =\frac{1}{2} \int {e^{2x}} \, d(2x) =\frac{e^{2x}}{2}$

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$\displaystyle = \frac{xe^{2x}}{2} - \int {\frac{e^{2x}}{2}} \, dx = \frac{xe^{2x}}{2} - \frac{1}{2} \int {e^{2x}} \, dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} +C$