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

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

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

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

2

Solution steps

\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

x \to - 5 + lim (\frac{- 2 x}{x + 5})

\int 10

x \to 1 - lim (x - 1)

y^{^{'}} = 1 - x + y - x y

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

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