integral of ((2x+1))/(x^2+4x-21)

解答

\int \frac{( 2 x + 1 )}{x ^{2} + 4 x - 21} d x

\frac{7}{10} ln ∣ x - 3 ∣ + \frac{13}{10} ln ∣ x + 7 ∣ + C

求解步骤

\int \frac{2 x + 1}{x ^{2} + 4 x - 21} d x

乘开

\frac{2 x + 1}{x ^{2} + 4 x - 21} : \frac{2 x}{x ^{2} + 4 x - 21} + \frac{1}{x ^{2} + 4 x - 21}

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{2 x}{x ^{2} + 4 x - 21} d x + \int \frac{1}{x ^{2} + 4 x - 21} d x

\int \frac{2 x}{x ^{2} + 4 x - 21} d x = 2 (\frac{3}{10} ln ∣ x - 3 ∣ + \frac{7}{10} ln ∣ x + 7 ∣)

\int \frac{1}{x ^{2} + 4 x - 21} d x = \frac{1}{10} ln ∣ x - 3 ∣ - \frac{1}{10} ln ∣ x + 7 ∣

= 2 (\frac{3}{10} ln ∣ x - 3 ∣ + \frac{7}{10} ln ∣ x + 7 ∣) + \frac{1}{10} ln ∣ x - 3 ∣ - \frac{1}{10} ln ∣ x + 7 ∣

化简

2 (\frac{3}{10} ln ∣ x - 3 ∣ + \frac{7}{10} ln ∣ x + 7 ∣) + \frac{1}{10} ln ∣ x - 3 ∣ - \frac{1}{10} ln ∣ x + 7 ∣ : \frac{7}{10} ln ∣ x - 3 ∣ + \frac{13}{10} ln ∣ x + 7 ∣

= \frac{7}{10} ln ∣ x - 3 ∣ + \frac{13}{10} ln ∣ x + 7 ∣

解答补常数

= \frac{7}{10} ln ∣ x - 3 ∣ + \frac{13}{10} ln ∣ x + 7 ∣ + C

\int (4 - x^{2})^{0.5} d x

\int sech (ln (x)) d x

\int - 2 + (x^{2} + 3 x) - x^{3} d x

\int \frac{5}{( 9 - x ^{2} ) ^{\frac{5}{2}}} d x

\int cos (x) (1 + sin (x))^{3} d x