integral of (x+1)[ln(x+1)+c]

解答

\int (x + 1) [ln (x + 1) + c] d x

\frac{1}{2} x^{2} ln (x + 1) - \frac{1}{4} x^{2} + \frac{c x ^{2}}{2} + c x - \frac{1}{2} x + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} + C

求解步骤

\int (x + 1) [ln (x + 1) + c] d x

乘开

(x + 1) [ln (x + 1) + c] : x ln (x + 1) + c x + ln (x + 1) + c

= \int x ln (x + 1) + c x + ln (x + 1) + c d x

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 x ln (x + 1) d x + \int c x d x + \int ln (x + 1) d x + \int c d x

\int x ln (x + 1) d x = \frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1

\int c x d x = \frac{c x ^{2}}{2}

\int ln (x + 1) d x = x ln (x + 1) + ln (x + 1) - x - 1

\int c d x = c x

= \frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1 + \frac{c x ^{2}}{2} + x ln (x + 1) + ln (x + 1) - x - 1 + c x

化简

\frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1 + \frac{c x ^{2}}{2} + x ln (x + 1) + ln (x + 1) - x - 1 + c x : \frac{1}{2} x^{2} ln (x + 1) - \frac{1}{4} x^{2} + \frac{c x ^{2}}{2} + c x - \frac{1}{2} x + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4}

= \frac{1}{2} x^{2} ln (x + 1) - \frac{1}{4} x^{2} + \frac{c x ^{2}}{2} + c x - \frac{1}{2} x + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4}

解答补常数

= \frac{1}{2} x^{2} ln (x + 1) - \frac{1}{4} x^{2} + \frac{c x ^{2}}{2} + c x - \frac{1}{2} x + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} + C

y^{2} y^{^{'}} = y^{2}

t an g e n t f (x) = x^{2} - 2 x

\int \frac{x ^{3}}{x ^{2} + 9} d x

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

a re a x = (y - 1)^{2}, x = 13 - y^{2}