integral of 19arctan(x)

解答

\int 19 arctan (x) d x

19 (x arctan (x) - \frac{1}{2} ln x^{2} + 1) + C

求解步骤

\int 19 arctan (x) d x

提出常数:

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

= 19 \cdot \int arctan (x) d x

使用分布积分法

= 19 (x arctan (x) - \int \frac{x}{x ^{2} + 1} d x)

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

= 19 (x arctan (x) - \frac{1}{2} ln x^{2} + 1)

解答补常数

= 19 (x arctan (x) - \frac{1}{2} ln x^{2} + 1) + C

f (x) = (4 x^{2} - 6 x)^{3}

d er i v a t i v e y = \frac{7 x ^{3} - 3 x ^{2} - 9}{x ^{8}}

\frac{d}{d x} (cos^{2} (x) \cdot sin (x))

\int_{0}^{4} 6 x d x

t an g e n t f (x) = x^{\frac{1}{2}}, (25, 5)