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

The integral of (2/x+1)^2 is -4/x+4ln|x|+x+C

integral of (2/x+1)^2

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

- \frac{4}{x} + 4 ln ∣ x ∣ + x + C

Solution steps

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

Expand

(\frac{2}{x} + 1)^{2} : \frac{4}{x ^{2}} + \frac{4}{x} + 1

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

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int 1 d x = x

= - \frac{4}{x} + 4 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{4}{x} + 4 ln ∣ x ∣ + x + C

\int a^{n y} d y

\int 2 x^{3} + 3 x^{2} - 10 x d x

\int \frac{x ^{2} + x + 8}{x ^{2} + 3 x - 4} d x

\int \frac{12 x ^{8} - x}{x ^{3}} d x

\int (e^{2 x} + 10)^{3} d x

What is the integral of (2/x+1)^2 ?
The integral of (2/x+1)^2 is -4/x+4ln|x|+x+C