integral of ((x^2+1))/((x-5)(x-4)^2)

解答

\int \frac{( x ^{2} + 1 )}{( x - 5 ) ( x - 4 ) ^{2}} d x

26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ + \frac{17}{x - 4} + C

求解步骤

\int \frac{x ^{2} + 1}{( x - 5 ) ( x - 4 ) ^{2}} d x

将

\frac{x ^{2} + 1}{( x - 5 ) ( x - 4 ) ^{2}}

用部份分式展开:

\frac{26}{x - 5} - \frac{25}{x - 4} - \frac{17}{( x - 4 ) ^{2}}

= \int \frac{26}{x - 5} - \frac{25}{x - 4} - \frac{17}{( x - 4 ) ^{2}} 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{26}{x - 5} d x - \int \frac{25}{x - 4} d x - \int \frac{17}{( x - 4 ) ^{2}} d x

\int \frac{26}{x - 5} d x = 26 ln ∣ x - 5 ∣

\int \frac{25}{x - 4} d x = 25 ln ∣ x - 4 ∣

\int \frac{17}{( x - 4 ) ^{2}} d x = - \frac{17}{x - 4}

= 26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ - (- \frac{17}{x - 4})

化简

= 26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ + \frac{17}{x - 4}

解答补常数

= 26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ + \frac{17}{x - 4} + C

\frac{d}{d x} (- \frac{1}{2 ( 1 + x ) ^{\frac{3}{2}}})

\int_{\frac{1}{4}}^{7} \frac{9}{3 4 x - 1} d x

d er i v a t i v e f (x) = - \frac{6}{( 1 + 3 x ) ^{2}}

\frac{d}{d x} (ln (\frac{( 2 )}{3 - x ^{2}}))

d er i v a t i v e \frac{3}{t ^{3}}