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

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

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

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

x^{4} + x^{3} + x^{2} + x + ln ∣ x ∣ - \frac{1}{2 x ^{2}} + C

Solution steps

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

\int 4 x^{3} d x = x^{4}

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= x^{4} + x^{3} + x^{2} + x + ln ∣ x ∣ - \frac{1}{2 x ^{2}}

Add a constant to the solution

= x^{4} + x^{3} + x^{2} + x + ln ∣ x ∣ - \frac{1}{2 x ^{2}} + C

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0

\int (\frac{1}{x + 1}) d x

\frac{d}{d x} (\frac{2 a}{x})

\frac{\partial}{\partial x} (7 e^{- 4 x})

\int_{2}^{6} f (x) d x

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