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

解答

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

\frac{1}{12} (x^{2} + 3)^{6} + C

求解步骤

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

使用换元积分法

= \int \frac{u ^{5}}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int u^{5} d u

使用幂法则

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

u = x^{2} + 3

代回

= \frac{1}{2} \cdot \frac{( x ^{2} + 3 ) ^{6}}{6}

化简

\frac{1}{2} \cdot \frac{( x ^{2} + 3 ) ^{6}}{6} : \frac{1}{12} (x^{2} + 3)^{6}

= \frac{1}{12} (x^{2} + 3)^{6}

解答补常数

= \frac{1}{12} (x^{2} + 3)^{6} + C

\frac{d}{d x} (3 x^{5} - 4 x^{4} + 9 x - 6)

\int \frac{sin ( x )}{2} d x

\frac{d}{d x} (sin (y) - y sin (x))

\int sec^{5} (2 x) tan (2 x) d x

x \to 1 + lim (\frac{6}{x - 1})