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

The integral from 1 to 2 of (x^3+1)/(x^2) is 2

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integral from 1 to 2 of (x^3+1)/(x^2)

\int_{1}^{2} \frac{x ^{3} + 1}{x ^{2}} d x

2

Solution steps

\int_{1}^{2} \frac{x ^{3} + 1}{x ^{2}} d x

Expand

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

= \int_{1}^{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_{1}^{2} x d x + \int_{1}^{2} \frac{1}{x ^{2}} d x

\int_{1}^{2} x d x = \frac{3}{2}

\int_{1}^{2} \frac{1}{x ^{2}} d x = \frac{1}{2}

= \frac{3}{2} + \frac{1}{2}

Simplify

\frac{3}{2} + \frac{1}{2} : 2

= 2

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What is the integral from 1 to 2 of (x^3+1)/(x^2) ?
The integral from 1 to 2 of (x^3+1)/(x^2) is 2