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受欢迎的微积分 >

integral of 1/(sqrt(x)(9x+4))

解答

\int \frac{1}{x ( 9 x + 4 )} d x

解答

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

求解步骤

\int \frac{1}{x ( 9 x + 4 )} d x

使用换元积分法

= \int \frac{2}{9 u ^{2} + 4} d u

提出常数:

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

= 2 \cdot \int \frac{1}{9 u ^{2} + 4} d u

使用换元积分法

= 2 \cdot \int \frac{1}{6 ( 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 \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

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

= 2 \cdot \frac{1}{6} arctan (v)

代回

= 2 \cdot \frac{1}{6} arctan (\frac{3}{2} x)

化简

2 \cdot \frac{1}{6} arctan (\frac{3}{2} x) : \frac{1}{3} arctan (\frac{3}{2} x)

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

解答补常数

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

作图

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