integral of (x^2+x-5)/(2x^3+3x^2-2x)

解答

\int \frac{x ^{2} + x - 5}{2 x ^{3} + 3 x ^{2} - 2 x} d x

\frac{5}{2} ln ∣ x ∣ - \frac{17}{10} ln ∣ 2 x - 1 ∣ - \frac{3}{10} ln ∣ x + 2 ∣ + C

求解步骤

\int \frac{x ^{2} + x - 5}{2 x ^{3} + 3 x ^{2} - 2 x} d x

将

\frac{x ^{2} + x - 5}{2 x ^{3} + 3 x ^{2} - 2 x}

用部份分式展开:

\frac{5}{2 x} - \frac{17}{5 ( 2 x - 1 )} - \frac{3}{10 ( x + 2 )}

= \int \frac{5}{2 x} - \frac{17}{5 ( 2 x - 1 )} - \frac{3}{10 ( x + 2 )} 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

= \int \frac{5}{2 x} d x - \int \frac{17}{5 ( 2 x - 1 )} d x - \int \frac{3}{10 ( x + 2 )} d x

\int \frac{5}{2 x} d x = \frac{5}{2} ln ∣ x ∣

\int \frac{17}{5 ( 2 x - 1 )} d x = \frac{17}{10} ln ∣ 2 x - 1 ∣

\int \frac{3}{10 ( x + 2 )} d x = \frac{3}{10} ln ∣ x + 2 ∣

= \frac{5}{2} ln ∣ x ∣ - \frac{17}{10} ln ∣ 2 x - 1 ∣ - \frac{3}{10} ln ∣ x + 2 ∣

解答补常数

= \frac{5}{2} ln ∣ x ∣ - \frac{17}{10} ln ∣ 2 x - 1 ∣ - \frac{3}{10} ln ∣ x + 2 ∣ + C

\int cos (ln (2 x)) d x

\int \frac{x}{x ^{2} - 2 x + 5} d x

\frac{d}{d x} (arccot (\frac{x}{4}))

\int \frac{y ^{3}}{( 2 + 3 y ^{4} ) ^{3}} d y

d er i v a t i v e sin (tan (3 x))