integral of x^3sqrt(15+x^4)

解答

\int x^{3} 15 + x^{4} d x

\frac{1}{6} (15 + x^{4})^{\frac{3}{2}} + C

求解步骤

\int x^{3} 15 + x^{4} 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 = 15 + x^{4}

代回

= \frac{1}{4} \cdot \frac{2}{3} (15 + x^{4})^{\frac{3}{2}}

化简

\frac{1}{4} \cdot \frac{2}{3} (15 + x^{4})^{\frac{3}{2}} : \frac{1}{6} (15 + x^{4})^{\frac{3}{2}}

= \frac{1}{6} (15 + x^{4})^{\frac{3}{2}}

解答补常数

= \frac{1}{6} (15 + x^{4})^{\frac{3}{2}} + C

\int \frac{1}{4 x ^{5}} d x

\int \frac{1}{- x ^{2} - x + 2} d x

\int ln^{2} (y) d y

\int x^{4} x d x

\int \frac{7 x}{e ^{2 x}} d x