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

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

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

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

\frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x - 6 ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{2 x ^{3} - 5 x ^{2} + x - 6}{x} : 2 x^{2} - 5 x + 1 - \frac{6}{x}

= \int 2 x^{2} - 5 x + 1 - \frac{6}{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 2 x^{2} d x - \int 5 x d x + \int 1 d x - \int \frac{6}{x} d x

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

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

\int 1 d x = x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

= \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x - 6 ln ∣ x ∣

Add a constant to the solution

= \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x - 6 ln ∣ x ∣ + C

x \to 0 lim (\frac{16}{x ^{4}})

f (x) = \frac{1}{x + 6}

\frac{e ^{y} + e ^{- y}}{2}

\int e^{y^{2}} d y

\int \frac{10}{8 x - 4 x ^{2}} d x

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