integral of 1/(25+16x^2)

解答

\int \frac{1}{25 + 16 x ^{2}} d x

\frac{1}{20} arctan (\frac{4}{5} x) + C

求解步骤

\int \frac{1}{25 + 16 x ^{2}} d x

使用换元积分法

= \int \frac{1}{20 ( 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}{20} \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 = arctan (u)

= \frac{1}{20} arctan (u)

u = \frac{4}{5} x

代回

= \frac{1}{20} arctan (\frac{4}{5} x)

解答补常数

= \frac{1}{20} arctan (\frac{4}{5} x) + C

\int_{- 1}^{1} (\frac{1}{x ^{2} + 1}) d x

x \to 0 - lim (\frac{1}{e ^{\frac{1}{x}} - 1})

d er i v a t i v e (x^{3} - x) (x^{2} + 2) (x^{2} + x - 1)

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

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