integral of (18x^2)/(x^4-41x^2+400)

解答

\int \frac{18 x ^{2}}{x ^{4} - 41 x ^{2} + 400} d x

4 ln ∣ x + 4 ∣ - 4 ln ∣ x - 4 ∣ - 5 ln ∣ x + 5 ∣ + 5 ln ∣ x - 5 ∣ + C

求解步骤

\int \frac{18 x ^{2}}{x ^{4} - 41 x ^{2} + 400} d x

提出常数:

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

= 18 \cdot \int \frac{x ^{2}}{x ^{4} - 41 x ^{2} + 400} d x

将

\frac{x ^{2}}{x ^{4} - 41 x ^{2} + 400}

用部份分式展开:

\frac{2}{9 ( x + 4 )} - \frac{2}{9 ( x - 4 )} - \frac{5}{18 ( x + 5 )} + \frac{5}{18 ( x - 5 )}

= 18 \cdot \int \frac{2}{9 ( x + 4 )} - \frac{2}{9 ( x - 4 )} - \frac{5}{18 ( x + 5 )} + \frac{5}{18 ( x - 5 )} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 18 (\int \frac{2}{9 ( x + 4 )} d x - \int \frac{2}{9 ( x - 4 )} d x - \int \frac{5}{18 ( x + 5 )} d x + \int \frac{5}{18 ( x - 5 )} d x)

\int \frac{2}{9 ( x + 4 )} d x = \frac{2}{9} ln ∣ x + 4 ∣

\int \frac{2}{9 ( x - 4 )} d x = \frac{2}{9} ln ∣ x - 4 ∣

\int \frac{5}{18 ( x + 5 )} d x = \frac{5}{18} ln ∣ x + 5 ∣

\int \frac{5}{18 ( x - 5 )} d x = \frac{5}{18} ln ∣ x - 5 ∣

= 18 (\frac{2}{9} ln ∣ x + 4 ∣ - \frac{2}{9} ln ∣ x - 4 ∣ - \frac{5}{18} ln ∣ x + 5 ∣ + \frac{5}{18} ln ∣ x - 5 ∣)

化简

18 (\frac{2}{9} ln ∣ x + 4 ∣ - \frac{2}{9} ln ∣ x - 4 ∣ - \frac{5}{18} ln ∣ x + 5 ∣ + \frac{5}{18} ln ∣ x - 5 ∣) : 4 ln ∣ x + 4 ∣ - 4 ln ∣ x - 4 ∣ - 5 ln ∣ x + 5 ∣ + 5 ln ∣ x - 5 ∣

= 4 ln ∣ x + 4 ∣ - 4 ln ∣ x - 4 ∣ - 5 ln ∣ x + 5 ∣ + 5 ln ∣ x - 5 ∣

解答补常数

= 4 ln ∣ x + 4 ∣ - 4 ln ∣ x - 4 ∣ - 5 ln ∣ x + 5 ∣ + 5 ln ∣ x - 5 ∣ + C

\frac{d}{d x} (4 x - x^{2} - 3)

y \to 15 lim (y + 3)

\int \frac{sin ( 2 e ^{- 4 x} )}{e ^{4 x}} d x

a re a f (x) = x^{3} - 3 x^{2} + 3 x, g (x) = x

d er i v a t i v e f (3) = 2 x^{2} + 4 x