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受欢迎的微积分 >

integral of 1/((2x+4)sqrt(2x-5))

解答

\int \frac{1}{( 2 x + 4 ) 2 x - 5} d x

解答

\frac{1}{3} arctan (\frac{2 x - 5}{3}) + C

求解步骤

\int \frac{1}{( 2 x + 4 ) 2 x - 5} d x

使用换元积分法

= \int \frac{1}{2 u u - 9} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{u u - 9} d u

使用换元积分法

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 9} d v

提出常数:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

使用换元积分法

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

提出常数:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

使用积分常见定则:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (w)

代回

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{2 x + 4 - 9}{3})

化简

\frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{2 x + 4 - 9}{3}) : \frac{1}{3} arctan (\frac{2 x - 5}{3})

= \frac{1}{3} arctan (\frac{2 x - 5}{3})

解答补常数

= \frac{1}{3} arctan (\frac{2 x - 5}{3}) + C

作图

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\int \frac{1}{6 u} d u

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

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\int (1 + ln (x)) 1 + (x ln (x))^{2} d x

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\int x \cdot 5^{- x^{2}} d x