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

The integral from 1 to 3 of pi(1/x+1)^2 is pi(2ln(3)+8/3)

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

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

π (2 ln (3) + \frac{8}{3})

Decimal

15.28036 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{1}^{3} (\frac{1}{x} + 1)^{2} d x

Expand

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

= π \cdot \int_{1}^{3} \frac{1}{x ^{2}} + \frac{2}{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

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

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

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

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

= π (\frac{2}{3} + 2 ln (3) + 2)

Combine the fractions

2 + \frac{2}{3} : \frac{8}{3}

= π (2 ln (3) + \frac{8}{3})

y^{2} e^{t} + (e^{t} + 2) y^{^{'}} = 0

\int \frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} d x

x \to \infty lim (0^{x})

x^{4} (e^{x})

4 sin^{2} (t)

What is the integral from 1 to 3 of pi(1/x+1)^2 ?
The integral from 1 to 3 of pi(1/x+1)^2 is pi(2ln(3)+8/3)