integral of 1/(xsqrt(9x-1))

解答

\int \frac{1}{x 9 x - 1} d x

2 arctan (9 x - 1) + C

求解步骤

\int \frac{1}{x 9 x - 1} d x

使用换元积分法

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

使用积分常见定则:

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

= 2 arctan (u)

u = 9 x - 1

代回

= 2 arctan (9 x - 1)

解答补常数

= 2 arctan (9 x - 1) + C

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