integral of (x(x^2-16)^2-1/(2x+5))

解答

\int (x (x^{2} - 16)^{2} - \frac{1}{2 x + 5}) d x

\frac{x ^{6}}{6} - 8 x^{4} + 128 x^{2} - \frac{1}{2} ln ∣ 2 x + 5 ∣ + C

求解步骤

\int x (x^{2} - 16)^{2} - \frac{1}{2 x + 5} 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 x (x^{2} - 16)^{2} d x - \int \frac{1}{2 x + 5} d x

\int x (x^{2} - 16)^{2} d x = \frac{x ^{6}}{6} - 8 x^{4} + 128 x^{2}

\int \frac{1}{2 x + 5} d x = \frac{1}{2} ln ∣ 2 x + 5 ∣

= \frac{x ^{6}}{6} - 8 x^{4} + 128 x^{2} - \frac{1}{2} ln ∣ 2 x + 5 ∣

解答补常数

= \frac{x ^{6}}{6} - 8 x^{4} + 128 x^{2} - \frac{1}{2} ln ∣ 2 x + 5 ∣ + C

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