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受欢迎的微积分 >

integral of sqrt((6t)^2+(6t^2)^2)

解答

\int (6 t)^{2} + (6 t^{2})^{2} d t

解答

2 (t^{2} + 1)^{\frac{3}{2}} + C

求解步骤

\int (6 t)^{2} + (6 t^{2})^{2} d t

化简

(6 t)^{2} + (6 t^{2})^{2} : 6 t^{2} 1 + t^{2}

= \int 6 t^{2} 1 + t^{2} d t

提出常数:

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

= 6 \cdot \int t^{2} 1 + t^{2} d t

t^{2} = (t),

假定

t \geq 0

= 6 \cdot \int t 1 + t^{2} d t

使用换元积分法

= 6 \cdot \int u^{2} d u

使用幂法则

= 6 \cdot \frac{u ^{3}}{3}

u = 1 + t^{2}

代回

= 6 \cdot \frac{( 1 + t ^{2} ) ^{3}}{3}

化简

6 \cdot \frac{( 1 + t ^{2} ) ^{3}}{3} : 2 (t^{2} + 1)^{\frac{3}{2}}

= 2 (t^{2} + 1)^{\frac{3}{2}}

解答补常数

= 2 (t^{2} + 1)^{\frac{3}{2}} + C

作图

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