integral from 1 to 2 of 1/(x^3+4x)

解答

\int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

\frac{- ln ( 2 ) + ln ( 5 )}{8}

十进制

0.11453 \dots

求解步骤

\int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

将

\frac{1}{x ^{3} + 4 x}

用部份分式展开:

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

= \int_{1}^{2} \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )} 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_{1}^{2} \frac{1}{4 x} d x - \int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x

\int_{1}^{2} \frac{1}{4 x} d x = \frac{1}{4} ln (2)

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= \frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5))

化简

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= \frac{- ln ( 2 ) + ln ( 5 )}{8}

l a pl a ce t r an s f or m (5 e^{9 t} - 4) e^{- 6 t}

l a pl a ce t r an s f or m (t)^{2}

\int \frac{4}{81 - ( x + 9 ) ^{2}} d x

y^{^{'}} + 7 y = 3

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