integral of (x^5+4)/(x^3+x^2)

解答

\int \frac{x ^{5} + 4}{x ^{3} + x ^{2}} d x

3 x + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} + 3 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣ - \frac{4}{x} + 3 + C

求解步骤

\int \frac{x ^{5} + 4}{x ^{3} + x ^{2}} d x

乘开

\frac{x ^{5} + 4}{x ^{3} + x ^{2}} : \frac{x ^{5}}{x ^{3} + x ^{2}} + \frac{4}{x ^{3} + x ^{2}}

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{x ^{5}}{x ^{3} + x ^{2}} d x + \int \frac{4}{x ^{3} + x ^{2}} d x

\int \frac{x ^{5}}{x ^{3} + x ^{2}} d x = \frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣

\int \frac{4}{x ^{3} + x ^{2}} d x = 4 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣)

= \frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣ + 4 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣)

化简

\frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣ + 4 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣) : 3 x + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} + 3 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣ - \frac{4}{x} + 3

= 3 x + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} + 3 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣ - \frac{4}{x} + 3

解答补常数

= 3 x + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} + 3 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣ - \frac{4}{x} + 3 + C

\int_{0}^{R} R^{2} - x^{2} d x

y^{^{'}} - 3 y = 0

\int 5 cos (x) + 4 sin (x) d x

\frac{\partial}{\partial y} ((x^{3} + (\frac{y}{2}))^{\frac{2}{3}})

\frac{d}{d x} (765480)