integral of 2sqrt((4-t^2)/(t^8))

解答

\int 2 \frac{4 - t ^{2}}{t ^{8}} d t

2 (- \frac{4 - t ^{2}}{3 t ^{3}} + \frac{4 - t ^{2}}{12 t}) + C

求解步骤

\int 2 \frac{4 - t ^{2}}{t ^{8}} d t

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{4 - t ^{2}}{t ^{8}} d t

化简

\frac{4 - t ^{2}}{t ^{8}} : \frac{4 - t ^{2}}{t ^{4}}

= 2 \cdot \int \frac{4 - t ^{2}}{t ^{4}} d t

使用分布积分法

= 2 (- \frac{4 - t ^{2}}{3 t ^{3}} - \int \frac{1}{3 t ^{2} - t ^{2} + 4} d t)

\int \frac{1}{3 t ^{2} - t ^{2} + 4} d t = - \frac{4 - t ^{2}}{12 t}

= 2 (- \frac{4 - t ^{2}}{3 t ^{3}} - (- \frac{4 - t ^{2}}{12 t}))

化简

= 2 (- \frac{4 - t ^{2}}{3 t ^{3}} + \frac{4 - t ^{2}}{12 t})

解答补常数

= 2 (- \frac{4 - t ^{2}}{3 t ^{3}} + \frac{4 - t ^{2}}{12 t}) + C

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