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

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

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

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

- \frac{1}{2 x ^{2}} - \frac{4}{x} + x + C

Solution steps

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

Simplify

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

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

Expand

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

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

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

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

\int 1 d x = x

= - \frac{1}{2 x ^{2}} - \frac{4}{x} + x

Add a constant to the solution

= - \frac{1}{2 x ^{2}} - \frac{4}{x} + x + C

\frac{37}{x ^{3}}, 1, 100

y^{^{'}} + 8 y = 1, y (0) = 1

f (t) = e^{2 t + 1}

\frac{d}{d x} (\frac{- 3 x + 8}{2 4 - x})

\frac{d}{d x} (\frac{x - 5}{4 x ^{2} e ^{x}})

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