integral of 4/((x-1)(x+3))

解答

\int \frac{4}{( x - 1 ) ( x + 3 )} d x

ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + C

求解步骤

\int \frac{4}{( x - 1 ) ( x + 3 )} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{( x - 1 ) ( x + 3 )} d x

将

\frac{1}{( x - 1 ) ( x + 3 )}

用部份分式展开:

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

= 4 \cdot \int \frac{1}{4 ( x - 1 )} - \frac{1}{4 ( x + 3 )} 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

= 4 (\int \frac{1}{4 ( x - 1 )} d x - \int \frac{1}{4 ( x + 3 )} d x)

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{4 ( x + 3 )} d x = \frac{1}{4} ln ∣ x + 3 ∣

= 4 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 3 ∣)

化简

4 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 3 ∣) : ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣

= ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣

解答补常数

= ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + C

t an g e n t f (x) = x^{2} x^{3} + 1, at x = 5

in v erse l a pl a ce \frac{4 s}{( s ^{2} + 16 )}

x \to 0 lim (18 - x^{2})

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

x \to \infty lim (\frac{8}{x !})