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受欢迎的微积分 >

integral of (e^xsqrt(e^x+2))/(e^x+4)

解答

\int \frac{e ^{x} e ^{x} + 2}{e ^{x} + 4} d x

解答

22 (- arctan (\frac{1}{2} (e^{x} + 2)) + \frac{1}{2} (e^{x} + 2)) + C

求解步骤

\int \frac{e ^{x} e ^{x} + 2}{e ^{x} + 4} d x

使用换元积分法

= \int \frac{u + 2}{u + 4} d u

使用换元积分法

= \int \frac{v}{v + 2} d v

使用换元积分法

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

使用换元积分法

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

提出常数:

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

= 22 \cdot \int \frac{t ^{2}}{t ^{2} + 1} d t

\frac{t ^{2}}{t ^{2} + 1} = - \frac{1}{t ^{2} + 1} + 1

= 22 \cdot \int - \frac{1}{t ^{2} + 1} + 1 d t

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

= 22 (- \int \frac{1}{t ^{2} + 1} d t + \int 1 d t)

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

\int 1 d t = t

= 22 (- arctan (t) + t)

代回

= 22 (- arctan (\frac{e ^{x} + 2}{2}) + \frac{e ^{x} + 2}{2})

化简

22 (- arctan (\frac{e ^{x} + 2}{2}) + \frac{e ^{x} + 2}{2}) : 22 (- arctan (\frac{1}{2} (e^{x} + 2)) + \frac{1}{2} (e^{x} + 2))

= 22 (- arctan (\frac{1}{2} (e^{x} + 2)) + \frac{1}{2} (e^{x} + 2))

解答补常数

= 22 (- arctan (\frac{1}{2} (e^{x} + 2)) + \frac{1}{2} (e^{x} + 2)) + C

作图

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