integral of 1/(4u^2+3)

解答

\int \frac{1}{4 u ^{2} + 3} d u

\frac{1}{2 3} arctan (\frac{2}{3} u) + C

求解步骤

\int \frac{1}{4 u ^{2} + 3} d u

使用换元积分法

= \int \frac{1}{2 3 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{2 3} \cdot \int \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2 3} arctan (v)

v = \frac{2}{3} u

代回

= \frac{1}{2 3} arctan (\frac{2}{3} u)

解答补常数

= \frac{1}{2 3} arctan (\frac{2}{3} u) + C

\int 8 arctan (2 y) d y

\frac{\partial}{\partial y} (e^{x^{2}} - y)

\int sin (120 π x) d x

t an g e n t f (x) = 4 x^{2} - x^{3}, (1, 3)

n or ma l y = - x^{2} - 2 x + 2, (- 3, - 1)