积分 ((1+x))/((1-x)^2)

解答

积分

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

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

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

= \frac{1}{1 - x} + ln ∣ 1 - x ∣ + \frac{1}{1 - x}

化简

\frac{1}{1 - x} + ln ∣ 1 - x ∣ + \frac{1}{1 - x} : \frac{2}{- x + 1} + ln ∣ - x + 1 ∣

= \frac{2}{- x + 1} + ln ∣ - x + 1 ∣

解答补常数

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

in t e g r a l \frac{7 x + 2}{1 - 25 x ^{2}}

in t e g r a l \frac{1}{( 5 + x ^{2} )}

in t e g r a l (\frac{1}{x - 1})^{\frac{- 1}{2}}

in t e g r a l \frac{x}{( x ^{2} - 1 ) ^{\frac{1}{2}}}

in t e g r a l (1 - x)^{2}