integral of (sqrt(u-1))/u

解答

\int \frac{u - 1}{u} d u

2 (- arctan (u - 1) + u - 1) + C

求解步骤

\int \frac{u - 1}{u} d u

使用换元积分法

= \int \frac{2 v ^{2}}{v ^{2} + 1} d v

提出常数:

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

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

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

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

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 \frac{1}{v ^{2} + 1} d v + \int 1 d v)

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= 2 (- arctan (v) + v)

v = u - 1

代回

= 2 (- arctan (u - 1) + u - 1)

解答补常数

= 2 (- arctan (u - 1) + u - 1) + C

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