integral of x^3(9+x^4)^6

解答

\int x^{3} (9 + x^{4})^{6} d x

\frac{1}{28} (9 + x^{4})^{7} + C

求解步骤

\int x^{3} (9 + x^{4})^{6} d x

使用换元积分法

= \int \frac{u ^{6}}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int u^{6} d u

使用幂法则

= \frac{1}{4} \cdot \frac{u ^{7}}{7}

u = 9 + x^{4}

代回

= \frac{1}{4} \cdot \frac{( 9 + x ^{4} ) ^{7}}{7}

化简

\frac{1}{4} \cdot \frac{( 9 + x ^{4} ) ^{7}}{7} : \frac{1}{28} (9 + x^{4})^{7}

= \frac{1}{28} (9 + x^{4})^{7}

解答补常数

= \frac{1}{28} (9 + x^{4})^{7} + C

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