What is the integral of (1/x+5)^2 ?

The integral of (1/x+5)^2 is 25x+10ln|x|-1/x+C

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integral of (1/x+5)^2

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

25 x + 10 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Multiply by the conjugate of

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

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

Simplify

(\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

Expand

\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

Apply the Sum Rule:

\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}

Add a constant to the solution

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

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What is the integral of (1/x+5)^2 ?
The integral of (1/x+5)^2 is 25x+10ln|x|-1/x+C