integral of x^2(6x^3+4)^5

解答

\int x^{2} (6 x^{3} + 4)^{5} d x

\frac{1}{108} (6 x^{3} + 4)^{6} + C

求解步骤

\int x^{2} (6 x^{3} + 4)^{5} d x

使用换元积分法

= \int \frac{u ^{5}}{18} d u

提出常数:

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

= \frac{1}{18} \cdot \int u^{5} d u

使用幂法则

= \frac{1}{18} \cdot \frac{u ^{6}}{6}

u = 6 x^{3} + 4

代回

= \frac{1}{18} \cdot \frac{( 6 x ^{3} + 4 ) ^{6}}{6}

化简

\frac{1}{18} \cdot \frac{( 6 x ^{3} + 4 ) ^{6}}{6} : \frac{1}{108} (6 x^{3} + 4)^{6}

= \frac{1}{108} (6 x^{3} + 4)^{6}

解答补常数

= \frac{1}{108} (6 x^{3} + 4)^{6} + C

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