zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of (e^tsqrt(e^t+2))/(e^t+6)

解答

\int \frac{e ^{t} e ^{t} + 2}{e ^{t} + 6} d t

解答

4 (- arctan (\frac{1}{2} e^{t} + 2) + \frac{1}{2} e^{t} + 2) + C

求解步骤

\int \frac{e ^{t} e ^{t} + 2}{e ^{t} + 6} d t

使用换元积分法

= \int \frac{u + 2}{u + 6} d u

使用换元积分法

= \int \frac{v}{v + 4} d v

使用换元积分法

= \int \frac{2 w ^{2}}{w ^{2} + 4} d w

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{w ^{2}}{w ^{2} + 4} d w

\frac{w ^{2}}{w ^{2} + 4} = - \frac{4}{w ^{2} + 4} + 1

= 2 \cdot \int - \frac{4}{w ^{2} + 4} + 1 d w

使用换元积分法

= 2 \cdot \int 2 (- \frac{1}{r ^{2} + 1} + 1) d r

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 2 \cdot \int - \frac{1}{r ^{2} + 1} + 1 d r

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (- \int \frac{1}{r ^{2} + 1} d r + \int 1 d r)

\int \frac{1}{r ^{2} + 1} d r = arctan (r)

\int 1 d r = r

= 2 \cdot 2 (- arctan (r) + r)

代回

= 2 \cdot 2 (- arctan (\frac{e ^{t} + 2}{2}) + \frac{e ^{t} + 2}{2})

化简

2 \cdot 2 (- arctan (\frac{e ^{t} + 2}{2}) + \frac{e ^{t} + 2}{2}) : 4 (- arctan (\frac{1}{2} e^{t} + 2) + \frac{1}{2} e^{t} + 2)

= 4 (- arctan (\frac{1}{2} e^{t} + 2) + \frac{1}{2} e^{t} + 2)

解答补常数

= 4 (- arctan (\frac{1}{2} e^{t} + 2) + \frac{1}{2} e^{t} + 2) + C

作图

流行的例子

y^{''}-1y^'-12y=0

y^{^{''}} - 1 y^{^{'}} - 12 y = 0

(\partial)/(\partial x)(xy-y+sin(1))

\frac{\partial}{\partial x} (x y - y + sin (1))

integral from 0 to 2 of 2^{-θ}

\int_{0}^{2} 2^{- θ} d θ

integral of sin^3(4x)sqrt(cos(4x))

\int sin^{3} (4 x) cos (4 x) d x

(\partial)/(\partial x)(x^{-1}y^2+xy^2+5xy)

\frac{\partial}{\partial x} (x^{- 1} y^{2} + x y^{2} + 5 x y)