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

The integral of 1/x (x+1)^2 is (x^2)/2+2x+ln|x|+C

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

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

\frac{x ^{2}}{2} + 2 x + ln ∣ x ∣ + C

Solution steps

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

Simplify

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

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

Expand

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

= \int x + 2 + \frac{1}{x} 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 x d x + \int 2 d x + \int \frac{1}{x} d x

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

\int 2 d x = 2 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} + 2 x + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + ln ∣ x ∣ + C

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