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

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

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

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2}} = x^{- 2}

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

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

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

= \frac{1}{2} + 1

Simplify

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

= \frac{3}{2}

sin (t + 1)

\int_{0}^{4} x^{2} - 4 x d x

\frac{1}{t + 1}

\int t e^{- 5 t} d t

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