integral of ((x+3))/(x^2+6x+1)

解答

\int \frac{( x + 3 )}{x ^{2} + 6 x + 1} d x

\frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1) + C

求解步骤

\int \frac{x + 3}{x ^{2} + 6 x + 1} d x

乘开

\frac{x + 3}{x ^{2} + 6 x + 1} : \frac{x}{x ^{2} + 6 x + 1} + \frac{3}{x ^{2} + 6 x + 1}

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{x}{x ^{2} + 6 x + 1} d x + \int \frac{3}{x ^{2} + 6 x + 1} d x

\int \frac{x}{x ^{2} + 6 x + 1} d x = \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22

\int \frac{3}{x ^{2} + 6 x + 1} d x = - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1)

= \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1)

解答补常数

= \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1) + C

\int x^{1.9} + 7 x^{2.5} d x

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

\int e^{\frac{t}{4}} d t

\int e^{- 100 x} d x

\int ln (0)