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

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

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

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

\frac{x ^{3}}{3} + 5 ln ∣ x ∣ + \frac{3}{x} + C

Solution steps

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

Expand

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

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

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

\int \frac{5}{x} d x = 5 ln ∣ x ∣

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

= \frac{x ^{3}}{3} + 5 ln ∣ x ∣ - (- \frac{3}{x})

Simplify

= \frac{x ^{3}}{3} + 5 ln ∣ x ∣ + \frac{3}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} + 5 ln ∣ x ∣ + \frac{3}{x} + C

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

\int x^{- 7} d x

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

in v erse l a pl a ce \frac{1}{( s ^{2} ( s + 1 ) )}

\frac{d}{d x} ((4 x^{2} - 13)^{- 11})

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