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受欢迎的微积分 >

integral of x(10x+10)^{100}

解答

\int x (10 x + 10)^{100} d x

解答

1 0^{100} (\frac{( x + 1 ) ^{102}}{102} - \frac{( x + 1 ) ^{101}}{101}) + C

求解步骤

\int x (10 x + 10)^{100} d x

分解

x (10 x + 10)^{100} : 1 0^{100} x (x + 1)^{100}

= \int 1 0^{100} x (x + 1)^{100} d x

提出常数:

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

= 1 0^{100} \cdot \int x (x + 1)^{100} d x

使用换元积分法

= 1 0^{100} \cdot \int (u - 1) u^{100} d u

乘开

(u - 1) u^{100} : u^{101} - u^{100}

= 1 0^{100} \cdot \int u^{101} - u^{100} 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

= 1 0^{100} (\int u^{101} d u - \int u^{100} d u)

\int u^{101} d u = \frac{u ^{102}}{102}

\int u^{100} d u = \frac{u ^{101}}{101}

= 1 0^{100} (\frac{u ^{102}}{102} - \frac{u ^{101}}{101})

u = x + 1

代回

= 1 0^{100} (\frac{( x + 1 ) ^{102}}{102} - \frac{( x + 1 ) ^{101}}{101})

解答补常数

= 1 0^{100} (\frac{( x + 1 ) ^{102}}{102} - \frac{( x + 1 ) ^{101}}{101}) + C

作图

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