integral of arctan(3x-1)

解答

\int arctan (3 x - 1) d x

\frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2) + C

求解步骤

\int arctan (3 x - 1) d x

使用换元积分法

= \int arctan (u) \frac{1}{3} d u

提出常数:

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

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

使用分布积分法

= \frac{1}{3} (u arctan (u) - \int \frac{u}{u ^{2} + 1} d u)

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

= \frac{1}{3} (u arctan (u) - \frac{1}{2} ln u^{2} + 1)

u = 3 x - 1

代回

= \frac{1}{3} ((3 x - 1) arctan (3 x - 1) - \frac{1}{2} ln (3 x - 1)^{2} + 1)

化简

\frac{1}{3} ((3 x - 1) arctan (3 x - 1) - \frac{1}{2} ln (3 x - 1)^{2} + 1) : \frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2)

= \frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2)

解答补常数

= \frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2) + C

\int (3 - x^{2}) \cdot e^{x} d x

\int \frac{2 x}{( x + 4 ) ( x - 5 )} d x

\int \frac{x ^{2} - 4 x + 1}{( x ^{2} - x + 1 ) ( x + 4 )} d x

\int sin^{7} (x) cos^{6} (x) d x

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