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

解答

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

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

求解步骤

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

乘开

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

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

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

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

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

解答补常数

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

\frac{\partial}{\partial x} (4 x (x^{2} + y^{2}) - 4 a^{2} x)

d er i v a t i v e \frac{a}{x}

in v erse l a pl a ce \frac{s}{( s - 1 )}

y (1 + x^{2}) y^{^{'}} - x (6 + y^{2}) = 0

d er i v a t i v e y = 5 x^{4} e^{x} + e^{x} x^{5}