What is the integral of (4x^3+3x^2+2x+1)/(x^2) ?

The integral of (4x^3+3x^2+2x+1)/(x^2) is 2x^2+3x+2ln|x|-1/x+C

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integral of (4x^3+3x^2+2x+1)/(x^2)

\int \frac{4 x ^{3} + 3 x ^{2} + 2 x + 1}{x ^{2}} d x

2 x^{2} + 3 x + 2 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

\int \frac{4 x ^{3} + 3 x ^{2} + 2 x + 1}{x ^{2}} d x

Expand

\frac{4 x ^{3} + 3 x ^{2} + 2 x + 1}{x ^{2}} : 4 x + 3 + \frac{2}{x} + \frac{1}{x ^{2}}

= \int 4 x + 3 + \frac{2}{x} + \frac{1}{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 4 x d x + \int 3 d x + \int \frac{2}{x} d x + \int \frac{1}{x ^{2}} d x

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

\int 3 d x = 3 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= 2 x^{2} + 3 x + 2 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= 2 x^{2} + 3 x + 2 ln ∣ x ∣ - \frac{1}{x} + C

\frac{d y}{d x} + x (y^{2} + y) = 0, y (2) = 1

2 e^{- 3 t}

\frac{11}{4} x^{2} d y - 14 d x = 0

\frac{d}{d x} ((x)^{7 x})

\int \frac{6 e ^{2 x}}{3 - e ^{4 x}} d x

What is the integral of (4x^3+3x^2+2x+1)/(x^2) ?
The integral of (4x^3+3x^2+2x+1)/(x^2) is 2x^2+3x+2ln|x|-1/x+C