integral of ((x^2-3x-7))/((2x+3)(x+1)^2)

解答

\int \frac{( x ^{2} - 3 x - 7 )}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

- \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ + \frac{3}{x + 1} + C

求解步骤

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

将

\frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 ) ^{2}}

用部份分式展开:

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

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

\int \frac{1}{2 x + 3} d x = \frac{1}{2} ln ∣ 2 x + 3 ∣

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

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

= - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ - (- \frac{3}{x + 1})

化简

= - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ + \frac{3}{x + 1}

解答补常数

= - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x + 1 ∣ + \frac{3}{x + 1} + C

d er i v a t i v e 3 x^{4} - 6 x^{3} + 1

d er i v a t i v e f (x) = \frac{x ^{2} + 8}{9 x - 2}

d er i v a t i v e x^{10} e^{x}

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

\frac{d}{d x} (ux)