What is the integral of (sqrt(x)-3/x)^2x ?

The integral of (sqrt(x)-3/x)^2x is 9ln|x|-4x^{3/2}+(x^3)/3+C

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integral of (sqrt(x)-3/x)^2x

\int (x - \frac{3}{x})^{2} x d x

9 ln ∣ x ∣ - 4 x^{\frac{3}{2}} + \frac{x ^{3}}{3} + C

Solution steps

\int (x - \frac{3}{x})^{2} x d x

Expand

(x - \frac{3}{x})^{2} x : \frac{9}{x} - 6 x^{\frac{1}{2}} + x^{2}

= \int \frac{9}{x} - 6 x^{\frac{1}{2}} + x^{2} 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{9}{x} d x - \int 6 x^{\frac{1}{2}} d x + \int x^{2} d x

\int \frac{9}{x} d x = 9 ln ∣ x ∣

\int 6 x^{\frac{1}{2}} d x = 4 x^{\frac{3}{2}}

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

= 9 ln ∣ x ∣ - 4 x^{\frac{3}{2}} + \frac{x ^{3}}{3}

Add a constant to the solution

= 9 ln ∣ x ∣ - 4 x^{\frac{3}{2}} + \frac{x ^{3}}{3} + C

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

\frac{d}{d y} (3 x^{2} y^{2} + \frac{x}{y})

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

- \frac{14}{x ^{3}}

4 x^{5} + 11

What is the integral of (sqrt(x)-3/x)^2x ?
The integral of (sqrt(x)-3/x)^2x is 9ln|x|-4x^{3/2}+(x^3)/3+C