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

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