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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{4 x ^{5}}{x ^{2}} d x = x^{4}

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

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

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

Add a constant to the solution

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

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