integral of 1/(49-x^2)

解答

\int \frac{1}{49 - x ^{2}} d x

\frac{1}{14} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1) + C

求解步骤

\int \frac{1}{49 - x ^{2}} d x

使用换元积分法

= \int \frac{1}{7 ( - u ^{2} + 1 )} d u

提出常数:

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

= \frac{1}{7} \cdot \int \frac{1}{- u ^{2} + 1} d u

使用积分常见定则:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{7} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

u = \frac{x}{7}

代回

= \frac{1}{7} (\frac{ln \frac{x}{7} + 1}{2} - \frac{ln \frac{x}{7} - 1}{2})

化简

\frac{1}{7} (\frac{ln \frac{x}{7} + 1}{2} - \frac{ln \frac{x}{7} - 1}{2}) : \frac{1}{14} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1)

= \frac{1}{14} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1)

解答补常数

= \frac{1}{14} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1) + C

d er i v a t i v e 3 3 x

\frac{\partial}{\partial x} ((x^{2} + y) e^{3 x + 4 y})

\frac{d y}{d x} = 3 y + y^{4}

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

\frac{d}{d y} (\frac{x}{x ^{2} + y ^{2}})