What is the integral from 2 to 3 of (4/(x^2)+1) ?

The integral from 2 to 3 of (4/(x^2)+1) is 5/3

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integral from 2 to 3 of (4/(x^2)+1)

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

\frac{5}{3}

Decimal Notation

1.66666 \dots

Solution steps

\int_{2}^{3} \frac{4}{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_{2}^{3} \frac{4}{x ^{2}} d x + \int_{2}^{3} 1 d x

\int_{2}^{3} \frac{4}{x ^{2}} d x = \frac{2}{3}

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

= \frac{2}{3} + 1

Simplify

\frac{2}{3} + 1 : \frac{5}{3}

= \frac{5}{3}

\int \frac{- 7}{( x ^{2} + 7 ) ^{\frac{3}{2}}} d x

n = 8 \sum \infty \frac{n}{ln ( n )}

e^{- t} sin (t)

x \to \infty lim (lo g_{e} (x^{2} - 1))

x \to 3 lim (f (x))

What is the integral from 2 to 3 of (4/(x^2)+1) ?
The integral from 2 to 3 of (4/(x^2)+1) is 5/3