What is the integral of (10x^3-5x^2+4x)/(x^2) ?

The integral of (10x^3-5x^2+4x)/(x^2) is 5x^2-5x+4ln|x|+C

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integral of (10x^3-5x^2+4x)/(x^2)

\int \frac{10 x ^{3} - 5 x ^{2} + 4 x}{x ^{2}} d x

5 x^{2} - 5 x + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{10 x ^{3} - 5 x ^{2} + 4 x}{x ^{2}} d x

Simplify

\frac{10 x ^{3} - 5 x ^{2} + 4 x}{x ^{2}} : \frac{10 x ^{2} - 5 x + 4}{x}

= \int \frac{10 x ^{2} - 5 x + 4}{x} d x

Expand

\frac{10 x ^{2} - 5 x + 4}{x} : 10 x - 5 + \frac{4}{x}

= \int 10 x - 5 + \frac{4}{x} 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 10 x d x - \int 5 d x + \int \frac{4}{x} d x

\int 10 x d x = 5 x^{2}

\int 5 d x = 5 x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= 5 x^{2} - 5 x + 4 ln ∣ x ∣

Add a constant to the solution

= 5 x^{2} - 5 x + 4 ln ∣ x ∣ + C

\frac{d}{d x} (tanh (2 x - 3))

f (x) = x^{2} + x, (1, 2)

(- 1, - 4), (0, 2)

- x^{2} + 3 x + 1, - x - 1, [0, 4]

\int \frac{- 3}{x} d x

What is the integral of (10x^3-5x^2+4x)/(x^2) ?
The integral of (10x^3-5x^2+4x)/(x^2) is 5x^2-5x+4ln|x|+C