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受欢迎的微积分 >

integral of 6/(9x^2+1)

解答

\int \frac{6}{9 x ^{2} + 1} d x

解答

2 arctan (3 x) + C

求解步骤

\int \frac{6}{9 x ^{2} + 1} d x

提出常数:

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

= 6 \cdot \int \frac{1}{9 x ^{2} + 1} d x

使用换元积分法

= 6 \cdot \int \frac{1}{3 ( u ^{2} + 1 )} d u

提出常数:

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

= 6 \cdot \frac{1}{3} \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)

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

u = 3 x

代回

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

化简

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

= 2 arctan (3 x)

解答补常数

= 2 arctan (3 x) + C

作图

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