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

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

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

\int (\frac{x ^{2} + 2 x + 1}{\frac{1}{x}}) d x

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

Solution steps

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

Simplify

= \int x (x^{2} + 2 x + 1) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} (x^{2} + 2 x + 1) - \int x^{2} (x + 1) d x

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

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

= \frac{1}{2} x^{2} (x^{2} + 2 x + 1) - (\frac{x ^{4}}{4} + \frac{x ^{3}}{3})

Simplify

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

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

Add a constant to the solution

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

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