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受欢迎的微积分 >

integral of (3sqrt(x))/(x+9)

解答

\int \frac{3 x}{x + 9} d x

解答

18 (- arctan (\frac{1}{3} x) + \frac{1}{3} x) + C

求解步骤

\int \frac{3 x}{x + 9} d x

提出常数:

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

= 3 \cdot \int \frac{x}{x + 9} d x

使用换元积分法

= 3 \cdot \int \frac{2 u ^{2}}{u ^{2} + 9} d u

提出常数:

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

= 3 \cdot 2 \cdot \int \frac{u ^{2}}{u ^{2} + 9} d u

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

= 3 \cdot 2 \cdot \int - \frac{9}{u ^{2} + 9} + 1 d u

使用换元积分法

= 3 \cdot 2 \cdot \int 3 (- \frac{1}{v ^{2} + 1} + 1) d v

提出常数:

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

= 3 \cdot 2 \cdot 3 \cdot \int - \frac{1}{v ^{2} + 1} + 1 d v

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

= 3 \cdot 2 \cdot 3 (- \int \frac{1}{v ^{2} + 1} d v + \int 1 d v)

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

\int 1 d v = v

= 3 \cdot 2 \cdot 3 (- arctan (v) + v)

代回

= 3 \cdot 2 \cdot 3 (- arctan (\frac{x}{3}) + \frac{x}{3})

化简

3 \cdot 2 \cdot 3 (- arctan (\frac{x}{3}) + \frac{x}{3}) : 18 (- arctan (\frac{1}{3} x) + \frac{1}{3} x)

= 18 (- arctan (\frac{1}{3} x) + \frac{1}{3} x)

解答补常数

= 18 (- arctan (\frac{1}{3} x) + \frac{1}{3} x) + C

作图

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x \to 6 lim (\frac{2}{( x - 6 ) ^{2}})

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tangent f(x)=sqrt(x),\at x=81

t an g e n t f (x) = x, at x = 81