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

解答

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

ln x^{2} - 2 x + 2 + C

求解步骤

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

乘开

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

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

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

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

化简

= ln x^{2} - 2 x + 2

解答补常数

= ln x^{2} - 2 x + 2 + C

\frac{d x}{d t} = 7 (x^{2} + 1), x (\frac{π}{4}) = 1

d er i v a t i v e 6 t^{2} - 4 t + 9

l a pl a ce t r an s f or m f (t) = 2 e^{- t} cos (4 t)

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

\frac{d}{d x} (12 x - x^{2})