What is the integral of ((6x^5-3x^2+4)/(x^2)) ?

The integral of ((6x^5-3x^2+4)/(x^2)) is (3x^4)/2-3x-4/x+C

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integral of ((6x^5-3x^2+4)/(x^2))

\int (\frac{6 x ^{5} - 3 x ^{2} + 4}{x ^{2}}) d x

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

Solution steps

\int \frac{6 x ^{5} - 3 x ^{2} + 4}{x ^{2}} d x

Expand

\frac{6 x ^{5} - 3 x ^{2} + 4}{x ^{2}} : 6 x^{3} - 3 + \frac{4}{x ^{2}}

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

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

\int 3 d x = 3 x

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{s i n (x)})

- \frac{2 d y}{d x} + 6 x^{2} = 0

\int 6 x^{2} (2 x^{3} - 5)^{5} d x

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

\frac{d}{d x} (e^{- x})

What is the integral of ((6x^5-3x^2+4)/(x^2)) ?
The integral of ((6x^5-3x^2+4)/(x^2)) is (3x^4)/2-3x-4/x+C