integral of ((1-x))/(x^2+1)

解答

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

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

求解步骤

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

乘开

\frac{1 - x}{x ^{2} + 1} : \frac{1}{x ^{2} + 1} - \frac{x}{x ^{2} + 1}

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

解答补常数

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

\frac{\partial}{\partial x} (x^{y + z})

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

x \to - 1 lim (x^{2} + 5 x + 6)

y^{^{''}} - 7 y^{^{'}} = - 3

\frac{\partial}{\partial x} (t^{3} e^{- \frac{x ^{2}}{2 t}})