integral of (7x^4-6x^2-x+5)/(x+2)

解答

\int \frac{7 x ^{4} - 6 x ^{2} - x + 5}{x + 2} d x

81 x^{2} + 123 x + 95 ln ∣ x + 2 ∣ + \frac{7}{4} (x + 2)^{4} - \frac{56}{3} (x + 2)^{3} - 78 + C

求解步骤

\int \frac{7 x ^{4} - 6 x ^{2} - x + 5}{x + 2} d x

乘开

\frac{7 x ^{4} - 6 x ^{2} - x + 5}{x + 2} : \frac{7 x ^{4}}{x + 2} - \frac{6 x ^{2}}{x + 2} - \frac{x}{x + 2} + \frac{5}{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{7 x ^{4}}{x + 2} d x - \int \frac{6 x ^{2}}{x + 2} d x - \int \frac{x}{x + 2} d x + \int \frac{5}{x + 2} d x

\int \frac{7 x ^{4}}{x + 2} d x = 7 (12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16)

\int \frac{6 x ^{2}}{x + 2} d x = 6 (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣)

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

\int \frac{5}{x + 2} d x = 5 ln ∣ x + 2 ∣

= 7 (12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16) - 6 (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣) - (x + 2 - 2 ln ∣ x + 2 ∣) + 5 ln ∣ x + 2 ∣

化简

7 (12 x^{2} + 16 x + 16 ln ∣ x + 2 ∣ + \frac{1}{4} (x + 2)^{4} - \frac{8}{3} (x + 2)^{3} - 16) - 6 (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣) - (x + 2 - 2 ln ∣ x + 2 ∣) + 5 ln ∣ x + 2 ∣ : 81 x^{2} + 123 x + 95 ln ∣ x + 2 ∣ + \frac{7}{4} (x + 2)^{4} - \frac{56}{3} (x + 2)^{3} - 78

= 81 x^{2} + 123 x + 95 ln ∣ x + 2 ∣ + \frac{7}{4} (x + 2)^{4} - \frac{56}{3} (x + 2)^{3} - 78

解答补常数

= 81 x^{2} + 123 x + 95 ln ∣ x + 2 ∣ + \frac{7}{4} (x + 2)^{4} - \frac{56}{3} (x + 2)^{3} - 78 + C

\frac{d}{d x} (sin^{2} (2 b x))

\frac{d}{d x} (ln ((2 x + 5)^{3}))

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

\int \frac{1}{tan ( x ) sec ( x )} d x

\frac{\partial}{\partial y} (x^{5} y^{10} cos (x y^{2}))