integral of 1/2 t^2e^{t^3}

解答

\int \frac{1}{2} t^{2} e^{t^{3}} d t

\frac{1}{6} e^{t^{3}} + C

求解步骤

\int \frac{1}{2} t^{2} e^{t^{3}} d t

提出常数:

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

= \frac{1}{2} \cdot \int t^{2} e^{t^{3}} d t

使用换元积分法

: \int \frac{e ^{u^{\frac{3}{2}}} u}{2 t} d u

= \frac{1}{2} \cdot \int \frac{e ^{u}}{3} d u

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int e^{u} d u

使用积分常见定则:

\int e^{u} d u = e^{u}

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

u = t^{3}

代回

= \frac{1}{2} \cdot \frac{1}{3} e^{t^{3}}

化简

\frac{1}{2} \cdot \frac{1}{3} e^{t^{3}} : \frac{1}{6} e^{t^{3}}

= \frac{1}{6} e^{t^{3}}

解答补常数

= \frac{1}{6} e^{t^{3}} + C

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

\int \frac{( 1 - x )}{9 - 4 x ^{2}} d x

\int sin (2 x) e^{s i n (x)} d x

\int 45 (1 + x^{2}) (7 + 15 x + 5 x^{3})^{16} d x

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