integral of (37x+5)/((6x+1)(x-1))

解答

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

\frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

求解步骤

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

将

\frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )}

用部份分式展开:

\frac{1}{6 x + 1} + \frac{6}{x - 1}

= \int \frac{1}{6 x + 1} + \frac{6}{x - 1} 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 \frac{1}{6 x + 1} d x + \int \frac{6}{x - 1} d x

\int \frac{1}{6 x + 1} d x = \frac{1}{6} ln ∣ 6 x + 1 ∣

\int \frac{6}{x - 1} d x = 6 ln ∣ x - 1 ∣

= \frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣

解答补常数

= \frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

d er i v a t i v e f (x) = x \cdot ln (x)

\frac{\partial}{\partial x} (x^{8} + 6 x y^{7})

d er i v a t i v e 6 + \frac{1}{x ^{2}}

y^{^{''}} + 4 y = 3 cos (2 t) - 5 sin (2 t)

t an g e n t y = x^{4} - 4 x^{2} + 10, at x = 1