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受欢迎的微积分 >

integral of (ln(1+sqrt(x)))/(1+sqrt(x))

解答

\int \frac{ln ( 1 + x )}{1 + x} d x

解答

2 ln (1 + x) + 2 x ln (1 + x) - 2 - 2 x - ln^{2} (1 + x) + C

求解步骤

\int \frac{ln ( 1 + x )}{1 + x} d x

使用换元积分法

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

提出常数:

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

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

乘开

\frac{( u - 1 ) ln ( u )}{u} : ln (u) - \frac{ln ( u )}{u}

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

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

\int ln (u) d u = u ln (u) - u

\int \frac{ln ( u )}{u} d u = \frac{ln ^{2} ( u )}{2}

= 2 (u ln (u) - u - \frac{ln ^{2} ( u )}{2})

u = 1 + x

代回

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

化简

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

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

解答补常数

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

作图

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