integral from 25 to x of sqrt(1+9t)

解答

\int_{25}^{x} 1 + 9 t d t

\frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 452 226}{27}

求解步骤

\int_{25}^{x} 1 + 9 t d t

使用换元积分法

= \int_{226}^{1 + 9 x} \frac{u}{9} d u

提出常数:

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

= \frac{1}{9} \cdot \int_{226}^{1 + 9 x} u d u

使用幂法则

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{226}^{1 + 9 x}

带入上下限计算后得:

\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{452 226}{3}

= \frac{1}{9} (\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{452 226}{3})

化简

= \frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 452 226}{27}

\frac{d}{d x} (6 x^{2} - 6 x)

\int_{1}^{3} 15 r^{3} ln (r) d r

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 3 cos (2 x)

\int (9 x)^{5} (9 x)^{3} + 1 d x

\int \frac{x ^{2} - 3}{( x ^{2} + 1 ) ( x - 1 )} d x