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

The integral of (1+1/x)^2 1/(x^2) is -1/x-1/(x^2)-1/(3x^3)+C

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

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

- \frac{1}{x} - \frac{1}{x ^{2}} - \frac{1}{3 x ^{3}} + C

Solution steps

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

Simplify

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

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

Expand

\frac{( x + 1 ) ^{2}}{x ^{4}} : \frac{1}{x ^{2}} + \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

= \int \frac{1}{x ^{2}} + \frac{2}{x ^{3}} + \frac{1}{x ^{4}} 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 ^{3}} d x + \int \frac{1}{x ^{4}} d x

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

\int \frac{2}{x ^{3}} d x = - \frac{1}{x ^{2}}

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

= - \frac{1}{x} - \frac{1}{x ^{2}} - \frac{1}{3 x ^{3}}

Add a constant to the solution

= - \frac{1}{x} - \frac{1}{x ^{2}} - \frac{1}{3 x ^{3}} + C

\frac{d}{d x} (arccos (3 x + 1))

\int 3^{x} x d x

f (0) = \frac{x}{x ^{2} + 4}

\int_{0}^{1} (\frac{x ^{3} + 1}{x + 1}) d x

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + y^{\frac{1}{2}})

What is the integral of (1+1/x)^2 1/(x^2) ?
The integral of (1+1/x)^2 1/(x^2) is -1/x-1/(x^2)-1/(3x^3)+C