integral of (1/x+5)^2

解答

\int (\frac{1}{x} + 5)^{2} d x

25 x + 10 ln ∣ x ∣ - \frac{1}{x} + C

求解步骤

\int (\frac{1}{x} + 5)^{2} d x

乘以

\frac{1}{x} + 5

的共轭根式

: \frac{\frac{1}{x ^{2}} - 25}{\frac{1}{x} - 5}

= \int (\frac{\frac{1}{x ^{2}} - 25}{\frac{1}{x} - 5})^{2} d x

化简

(\frac{\frac{1}{x ^{2}} - 25}{\frac{1}{x} - 5})^{2} : \frac{( 5 x + 1 ) ^{2}}{x ^{2}}

= \int \frac{( 5 x + 1 ) ^{2}}{x ^{2}} d x

乘开

\frac{( 5 x + 1 ) ^{2}}{x ^{2}} : 25 + \frac{10}{x} + \frac{1}{x ^{2}}

= \int 25 + \frac{10}{x} + \frac{1}{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 25 d x + \int \frac{10}{x} d x + \int \frac{1}{x ^{2}} d x

\int 25 d x = 25 x

\int \frac{10}{x} d x = 10 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= 25 x + 10 ln ∣ x ∣ - \frac{1}{x}

解答补常数

= 25 x + 10 ln ∣ x ∣ - \frac{1}{x} + C