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受欢迎的微积分 >

integral of (10^x-1)^2

解答

\int (1 0^{x} - 1)^{2} d x

解答

\frac{2 5 ^{x} \cdot 2 ^{2 x - 1} - 5 ^{x} \cdot 2 ^{1 + x} + ln ( 10 ) x}{ln ( 10 )} + C

求解步骤

\int (1 0^{x} - 1)^{2} d x

因式分解

1 0^{x} - 1 : - (- 1 0^{x} + 1)

= \int (- (- 1 0^{x} + 1))^{2} d x

化简

(- (- 1 0^{x} + 1))^{2} : (- 1 0^{x} + 1)^{2}

= \int (- 1 0^{x} + 1)^{2} d x

使用换元积分法

= \int \frac{( - u + 1 ) ^{2}}{ln ( 10 ) u} d u

提出常数:

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

= \frac{1}{ln ( 10 )} \cdot \int \frac{( - u + 1 ) ^{2}}{u} d u

乘开

\frac{( - u + 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \frac{1}{ln ( 10 )} \cdot \int u - 2 + \frac{1}{u} 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

= \frac{1}{ln ( 10 )} (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{ln ( 10 )} (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

u = 1 0^{x}

代回

= \frac{1}{ln ( 10 )} (\frac{( 1 0 ^{x} ) ^{2}}{2} - 2 \cdot 1 0^{x} + ln ∣ 1 0^{x} ∣)

化简

\frac{1}{ln ( 10 )} (\frac{( 1 0 ^{x} ) ^{2}}{2} - 2 \cdot 1 0^{x} + ln ∣ 1 0^{x} ∣) : \frac{2 5 ^{x} \cdot 2 ^{2 x - 1} - 5 ^{x} \cdot 2 ^{1 + x} + ln ( 10 ) x}{ln ( 10 )}

= \frac{2 5 ^{x} \cdot 2 ^{2 x - 1} - 5 ^{x} \cdot 2 ^{1 + x} + ln ( 10 ) x}{ln ( 10 )}

解答补常数

= \frac{2 5 ^{x} \cdot 2 ^{2 x - 1} - 5 ^{x} \cdot 2 ^{1 + x} + ln ( 10 ) x}{ln ( 10 )} + C

作图

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