What is the integral from 0 to 5 of 2pix(1/(x+4)) ?

The integral from 0 to 5 of 2pix(1/(x+4)) is 2pi(5-4(2ln(3)-2ln(2)))

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integral from 0 to 5 of 2pix(1/(x+4))

\int_{0}^{5} 2 π x (\frac{1}{x + 4}) d x

2 π (5 - 4 (2 ln (3) - 2 ln (2)))

Decimal

11.03502 \dots

Solution steps

\int_{0}^{5} 2 π x \frac{1}{x + 4} d x

Take the constant out:

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

= 2 π \cdot \int_{0}^{5} x \frac{1}{x + 4} d x

Simplify

x \frac{1}{x + 4} : \frac{x}{x + 4}

= 2 π \cdot \int_{0}^{5} \frac{x}{x + 4} d x

Apply u-substitution

: \int_{4}^{9} \frac{u - 4}{u} d u

= 2 π \cdot \int_{4}^{9} \frac{u - 4}{u} d u

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

= 2 π \cdot \int_{4}^{9} 1 - \frac{4}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 π (\int_{4}^{9} 1 d u - \int_{4}^{9} \frac{4}{u} d u)

\int_{4}^{9} 1 d u = 5

\int_{4}^{9} \frac{4}{u} d u = 4 (2 ln (3) - 2 ln (2))

= 2 π (5 - 4 (2 ln (3) - 2 ln (2)))

- 1.621 s - 0.462945095

\frac{\partial}{\partial φ} (ρ sin (φ) cos (θ))

y^{^{''}} - 4 y^{^{'}} + 5 y = 2020 e^{- t}

5 2 x^{3} + 8 x (2.2)

(- 7.1) (- 7.9)

What is the integral from 0 to 5 of 2pix(1/(x+4)) ?
The integral from 0 to 5 of 2pix(1/(x+4)) is 2pi(5-4(2ln(3)-2ln(2)))