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

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

Popular >

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

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

- \frac{1}{x} + 2 ln ∣ x ∣ + x + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 1 d x = x

= - \frac{1}{x} + 2 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{1}{x} + 2 ln ∣ x ∣ + x + C

\frac{d y}{d x} = \frac{1}{y ^{2}}

f (x) = 3 x - 12 x^{3}, at x = 0

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

f (x) = 4 x^{2} e^{2 x^{2} + x - 1}

\int 13 - (\frac{x}{40}) d x

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