关于含lnx的不定积分案例

∫ ln ⁡ x x 3 d x \int \frac{\ln x}{x^3} dx x3lnxdx 进行积分:

u = ln ⁡ x u = \ln x u=lnx d v = 1 x 3 d x dv = \frac{1}{x^3} dx dv=x31dx,则 d u = 1 x d x du = \frac{1}{x} dx du=x1dx v = − 1 2 x 2 v = -\frac{1}{2x^2} v=2x21。则有:
∫ ln ⁡ x x 3 d x = − ln ⁡ x 2 x 2 + ∫ 1 2 x 3 d x \int \frac{\ln x}{x^3} dx = -\frac{\ln x}{2x^2} + \int \frac{1}{2x^3} dx x3lnxdx=2x2lnx+2x31dx
对于第二个积分,我们可以再次使用分部积分法,设 u = 1 2 x 2 u = \frac{1}{2x^2} u=2x21 d v = 1 x 3 d x dv = \frac{1}{x^3}dx dv=x31dx,则 d u = − 1 x 3 d x du = -\frac{1}{x^3}dx du=x31dx v = − 1 2 x 2 v = -\frac{1}{2x^2} v=2x21。则有:
∫ 1 2 x 3 d x = − 1 4 x 2 \int \frac{1}{2x^3} dx = -\frac{1}{4x^2} 2x31dx=4x21
将上述结果带回原式,得到:
∫ ln ⁡ x x 3 d x = − ln ⁡ x 2 x 2 − 1 4 x 2 + C \int \frac{\ln x}{x^3} dx = -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C x3lnxdx=2x2lnx4x21+C
其中 C C C 为任意常数。原积分的结果为 − ln ⁡ x 2 x 2 − 1 4 x 2 + C -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C 2x2lnx4x21+C

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