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受欢迎的微积分 >

integral of 7/((2x+1)[1+ln(2x+1)]^2)

解答

\int \frac{7}{( 2 x + 1 ) [ 1 + ln ( 2 x + 1 ) ] ^{2}} d x

解答

- \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )} + C

求解步骤

\int \frac{7}{( 2 x + 1 ) [ 1 + ln ( 2 x + 1 ) ] ^{2}} d x

提出常数:

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

= 7 \cdot \int \frac{1}{( 2 x + 1 ) [ 1 + ln ( 2 x + 1 ) ] ^{2}} d x

化简

(1 + ln (2 x + 1))^{2} : (1 + ln (2 x + 1)) (1 + ln (2 x + 1))

= 7 \cdot \int \frac{1}{( 2 x + 1 ) ( 1 + ln ( 2 x + 1 ) ) ( 1 + ln ( 2 x + 1 ) )} d x

(2 x + 1) (1 + ln (2 x + 1)) (1 + ln (2 x + 1)) = (2 x + 1) (1 + ln (2 x + 1))^{2}

= 7 \cdot \int \frac{1}{( 2 x + 1 ) ( 1 + ln ( 2 x + 1 ) ) ^{2}} d x

使用换元积分法

= 7 \cdot \int \frac{1}{2 u ( 1 + ln ( u ) ) ^{2}} d u

提出常数:

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

= 7 \cdot \frac{1}{2} \cdot \int \frac{1}{u ( 1 + ln ( u ) ) ^{2}} d u

使用换元积分法

= 7 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

使用幂法则

= 7 \cdot \frac{1}{2} (- \frac{1}{v})

代回

= 7 \cdot \frac{1}{2} (- \frac{1}{1 + ln ( 2 x + 1 )})

化简

7 \cdot \frac{1}{2} (- \frac{1}{1 + ln ( 2 x + 1 )}) : - \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )}

= - \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )}

解答补常数

= - \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )} + C

作图

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