integral of sqrt(x^4-1)x^3

解答

\int x^{4} - 1 x^{3} d x

\frac{1}{6} (x^{4} - 1)^{\frac{3}{2}} + C

求解步骤

\int x^{4} - 1 x^{3} d x

使用换元积分法

= \int \frac{u}{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 d u

使用幂法则

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

u = x^{4} - 1

代回

= \frac{1}{4} \cdot \frac{2}{3} (x^{4} - 1)^{\frac{3}{2}}

化简

\frac{1}{4} \cdot \frac{2}{3} (x^{4} - 1)^{\frac{3}{2}} : \frac{1}{6} (x^{4} - 1)^{\frac{3}{2}}

= \frac{1}{6} (x^{4} - 1)^{\frac{3}{2}}

解答补常数

= \frac{1}{6} (x^{4} - 1)^{\frac{3}{2}} + C

\int 7 x^{3} + 3 3 x^{2} d x

\int e^{c o s (y)} sin (2 y) d y

\int \frac{2 x ^{2} + 3 x + 2}{x ^{3} + 4 x ^{2} + 6 x + 4} d x

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

\int (x 8 - x^{2}) d x