What is the integral of ((3+5x-6x^2-7x^3)/(2x^2)) ?

The integral of ((3+5x-6x^2-7x^3)/(2x^2)) is 1/2 (-3/x+5ln|x|-6x-(7x^2)/2)+C

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integral of ((3+5x-6x^2-7x^3)/(2x^2))

\int (\frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{2 x ^{2}}) d x

\frac{1}{2} (- \frac{3}{x} + 5 ln ∣ x ∣ - 6 x - \frac{7 x ^{2}}{2}) + C

Solution steps

\int \frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{2 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{x ^{2}} d x

Expand

\frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{x ^{2}} : \frac{3}{x ^{2}} + \frac{5}{x} - 6 - 7 x

= \frac{1}{2} \cdot \int \frac{3}{x ^{2}} + \frac{5}{x} - 6 - 7 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

= \frac{1}{2} (\int \frac{3}{x ^{2}} d x + \int \frac{5}{x} d x - \int 6 d x - \int 7 x d x)

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

\int \frac{5}{x} d x = 5 ln ∣ x ∣

\int 6 d x = 6 x

\int 7 x d x = \frac{7 x ^{2}}{2}

= \frac{1}{2} (- \frac{3}{x} + 5 ln ∣ x ∣ - 6 x - \frac{7 x ^{2}}{2})

Add a constant to the solution

= \frac{1}{2} (- \frac{3}{x} + 5 ln ∣ x ∣ - 6 x - \frac{7 x ^{2}}{2}) + C

f (x) = x^{2} - 5 x + 5, at x = 4

2 x^{3} + 3 x^{2} - 12 x + 4

cos (6 x)

\frac{\partial}{\partial z} (z x^{2} + y^{2})

\frac{\partial}{\partial y} (x^{2} + 2 x y)

What is the integral of ((3+5x-6x^2-7x^3)/(2x^2)) ?
The integral of ((3+5x-6x^2-7x^3)/(2x^2)) is 1/2 (-3/x+5ln|x|-6x-(7x^2)/2)+C