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受欢迎的微积分 >

integral of (5x+3)\sqrt[3]{5x^2+6x}

解答

\int (5 x + 3) 3 5 x^{2} + 6 x d x

解答

\frac{3}{40 3 5} (25 x^{2} + 30 x)^{\frac{4}{3}} + C

求解步骤

\int (5 x + 3) 3 5 x^{2} + 6 x d x

使用换元积分法

= \int \frac{u 3 u ^{2} - 9}{5 3 5} d u

提出常数:

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

= \frac{1}{5 3 5} \cdot \int u 3 u^{2} - 9 d u

使用换元积分法

= \frac{1}{5 3 5} \cdot \int \frac{3 v}{2} d v

提出常数:

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

= \frac{1}{5 3 5} \cdot \frac{1}{2} \cdot \int 3 v d v

使用幂法则

= \frac{1}{5 3 5} \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

代回

= \frac{1}{5 3 5} \cdot \frac{1}{2} \cdot \frac{3}{4} ((5 x + 3)^{2} - 9)^{\frac{4}{3}}

化简

\frac{1}{5 3 5} \cdot \frac{1}{2} \cdot \frac{3}{4} ((5 x + 3)^{2} - 9)^{\frac{4}{3}} : \frac{3}{40 3 5} (25 x^{2} + 30 x)^{\frac{4}{3}}

= \frac{3}{40 3 5} (25 x^{2} + 30 x)^{\frac{4}{3}}

解答补常数

= \frac{3}{40 3 5} (25 x^{2} + 30 x)^{\frac{4}{3}} + C

作图

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y^{''}+7y^'+10y=4te^{4t}

y^{^{''}} + 7 y^{^{'}} + 10 y = 4 t e^{4 t}

sum from n=1 to infinity of (cos(1.4))^n

n = 1 \sum \infty (cos (1.4))^{n}

integral from 0 to t of t^2e^{-5t}

\int_{0}^{t} t^{2} e^{- 5 t} d t

d/(dt)(e^{-6t})

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y^'-6+y/((4(x+250)))=0,y(0)=70

y^{^{'}} - 6 + \frac{y}{( 4 ( x + 250 ) )} = 0, y (0) = 70