integral of (x^2)/((1+x^3)^{1.9)}

解答

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.9}} d x

- \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}} + C

求解步骤

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.9}} d x

使用换元积分法

= \int \frac{1}{3 u ^{1.9}} d u

提出常数:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{1.9}} d u

使用幂法则

= \frac{1}{3} (- \frac{1.11111 \dots}{u ^{0.9}})

u = 1 + x^{3}

代回

= \frac{1}{3} (- \frac{1.11111 \dots}{( 1 + x ^{3} ) ^{0.9}})

化简

\frac{1}{3} (- \frac{1.11111 \dots}{( 1 + x ^{3} ) ^{0.9}}) : - \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}}

= - \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}}

解答补常数

= - \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}} + C

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