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

解答

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

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

求解步骤

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

乘开

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

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

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

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

化简

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

解答补常数

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

in v erse l a pl a ce \frac{1}{s ^{\frac{3}{2}}}

6 y^{^{''}} + 31 y = 0, y (0) = 6, y^{^{'}} (0) = 2

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

t an g e n t y = \frac{1}{x ^{2}}, (3, \frac{1}{9})

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}