integral from 6 to 8 of y/(y^2-2y-3)

解答

\int_{6}^{8} \frac{y}{y ^{2} - 2 y - 3} d y

\frac{- ln ( 3 ) - ln ( 7 ) + 3 ln ( 5 )}{4}

十进制

0.44594 \dots

求解步骤

\int_{6}^{8} \frac{y}{y ^{2} - 2 y - 3} d y

将

\frac{y}{y ^{2} - 2 y - 3}

用部份分式展开:

\frac{1}{4 ( y + 1 )} + \frac{3}{4 ( y - 3 )}

= \int_{6}^{8} \frac{1}{4 ( y + 1 )} + \frac{3}{4 ( y - 3 )} d y

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_{6}^{8} \frac{1}{4 ( y + 1 )} d y + \int_{6}^{8} \frac{3}{4 ( y - 3 )} d y

\int_{6}^{8} \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} (2 ln (3) - ln (7))

\int_{6}^{8} \frac{3}{4 ( y - 3 )} d y = \frac{3}{4} (ln (5) - ln (3))

= \frac{1}{4} (2 ln (3) - ln (7)) + \frac{3}{4} (ln (5) - ln (3))

化简

\frac{1}{4} (2 ln (3) - ln (7)) + \frac{3}{4} (ln (5) - ln (3)) : \frac{- ln ( 3 ) - ln ( 7 ) + 3 ln ( 5 )}{4}

= \frac{- ln ( 3 ) - ln ( 7 ) + 3 ln ( 5 )}{4}

\int (x^{3} - 1)^{4} x^{2} d x

n = 1 \sum \infty \frac{1}{n ( n + 2 )}

t an g e n t 1 - 4 x

n = 1 \sum \infty \frac{( 1 )}{n ^{2}}

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