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受欢迎的微积分 >

integral of ((16x^2))/((sqrt(2x+5))^5)

解答

\int \frac{( 16 x ^{2} )}{( 2 x + 5 ) ^{5}} d x

解答

2 (\frac{20}{2 x + 5} + 2 2 x + 5 - \frac{50}{3 ( 2 x + 5 ) ^{\frac{3}{2}}}) + C

求解步骤

\int \frac{16 x ^{2}}{( 2 x + 5 ) ^{5}} d x

提出常数:

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

= 16 \cdot \int \frac{x ^{2}}{( 2 x + 5 ) ^{5}} d x

使用换元积分法

= 16 \cdot \int \frac{( u - 5 ) ^{2}}{8 u ^{\frac{5}{2}}} d u

提出常数:

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

= 16 \cdot \frac{1}{8} \cdot \int \frac{( u - 5 ) ^{2}}{u ^{\frac{5}{2}}} d u

乘开

\frac{( u - 5 ) ^{2}}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{10}{u ^{\frac{3}{2}}} + \frac{25}{u ^{\frac{5}{2}}}

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{10}{u ^{\frac{3}{2}}} + \frac{25}{u ^{\frac{5}{2}}} d u

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

= 16 \cdot \frac{1}{8} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{10}{u ^{\frac{3}{2}}} d u + \int \frac{25}{u ^{\frac{5}{2}}} d u)

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{10}{u ^{\frac{3}{2}}} d u = - \frac{20}{u}

\int \frac{25}{u ^{\frac{5}{2}}} d u = - \frac{50}{3 u ^{\frac{3}{2}}}

= 16 \cdot \frac{1}{8} (2 u^{\frac{1}{2}} - (- \frac{20}{u}) - \frac{50}{3 u ^{\frac{3}{2}}})

u = 2 x + 5

代回

= 16 \cdot \frac{1}{8} (2 (2 x + 5)^{\frac{1}{2}} - (- \frac{20}{2 x + 5}) - \frac{50}{3 ( 2 x + 5 ) ^{\frac{3}{2}}})

化简

16 \cdot \frac{1}{8} (2 (2 x + 5)^{\frac{1}{2}} - (- \frac{20}{2 x + 5}) - \frac{50}{3 ( 2 x + 5 ) ^{\frac{3}{2}}}) : 2 (\frac{20}{2 x + 5} + 2 2 x + 5 - \frac{50}{3 ( 2 x + 5 ) ^{\frac{3}{2}}})

= 2 (\frac{20}{2 x + 5} + 2 2 x + 5 - \frac{50}{3 ( 2 x + 5 ) ^{\frac{3}{2}}})

解答补常数

= 2 (\frac{20}{2 x + 5} + 2 2 x + 5 - \frac{50}{3 ( 2 x + 5 ) ^{\frac{3}{2}}}) + C

作图

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