integral of (x^2)/(x-4)

解答

\int \frac{x ^{2}}{x - 4} d x

\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣ + C

求解步骤

\int \frac{x ^{2}}{x - 4} d x

使用换元积分法

= \int \frac{( u + 4 ) ^{2}}{u} d u

乘开

\frac{( u + 4 ) ^{2}}{u} : u + 8 + \frac{16}{u}

= \int u + 8 + \frac{16}{u} d u

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 u d u + \int 8 d u + \int \frac{16}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 8 d u = 8 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{u ^{2}}{2} + 8 u + 16 ln ∣ u ∣

u = x - 4

代回

= \frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣

解答补常数

= \frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣ + C

d er i v a t i v e 5

\frac{d y}{d x} + 6 \cdot (tan (6 x)) \cdot y = 2 \cdot cos (6 x)

\int \frac{33 x ^{2}}{x ^{2} - 4} d x

\int_{0}^{π} x cos^{2} (x) d x

x \to 0 lim ((1 - 7 x)^{\frac{1}{x}})