integral of 1/(4+49x^2)

解答

\int \frac{1}{4 + 49 x ^{2}} d x

\frac{1}{14} arctan (\frac{7}{2} x) + C

求解步骤

\int \frac{1}{4 + 49 x ^{2}} d x

使用换元积分法

= \int \frac{1}{14 ( u ^{2} + 1 )} d u

提出常数:

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

= \frac{1}{14} \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)

= \frac{1}{14} arctan (u)

u = \frac{7}{2} x

代回

= \frac{1}{14} arctan (\frac{7}{2} x)

解答补常数

= \frac{1}{14} arctan (\frac{7}{2} x) + C

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