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

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

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

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

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

Solution steps

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

Simplify

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

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int 1 d x = x

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

Add a constant to the solution

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

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