integral of arctan(3t)

解答

\int arctan (3 t) d t

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

求解步骤

\int arctan (3 t) d t

使用分布积分法

= t arctan (3 t) - \int \frac{3 t}{9 t ^{2} + 1} d t

\int \frac{3 t}{9 t ^{2} + 1} d t = \frac{1}{6} ln 9 t^{2} + 1

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

解答补常数

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

\int \frac{1 - x}{( 1 + x ) 3 x} d x

\int \frac{e ^{4 v}}{( 1 + 3 e ^{4 v} ) ^{2}} d v

\int 2 x cos (\frac{x}{2}) d x

\int (\frac{sin ( x )}{1 + tan ( x )})^{2} d x

\int csc^{2} (2 a x) d x