What is the integral from 0 to 4 of sqrt(x(x+2)) ?

The integral from 0 to 4 of sqrt(x(x+2)) is -1/2 ln(2sqrt(6)+5)+5sqrt(6)

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integral from 0 to 4 of sqrt(x(x+2))

\int_{0}^{4} x (x + 2) d x

- \frac{1}{2} ln (26 + 5) + 56

Decimal

11.10123 \dots

Solution steps

\int_{0}^{4} x (x + 2) d x

Complete the square

x (x + 2) : (x + 1)^{2} - 1

= \int_{0}^{4} (x + 1)^{2} - 1 d x

Apply u-substitution

: \int_{1}^{5} u^{2} - 1 d u

= \int_{1}^{5} u^{2} - 1 d u

Apply Trigonometric Substitution

: \int_{0}^{arcsec (5)} tan^{2} (v) sec (v) d v

= \int_{0}^{arcsec (5)} tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \int_{0}^{arcsec (5)} (- 1 + sec^{2} (v)) sec (v) d v

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= \int_{0}^{arcsec (5)} - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

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

= - \int_{0}^{arcsec (5)} sec (v) d v + \int_{0}^{arcsec (5)} sec^{3} (v) d v

\int_{0}^{arcsec (5)} sec (v) d v = ln (26 + 5)

\int_{0}^{arcsec (5)} sec^{3} (v) d v = 56 + \frac{1}{2} ln (26 + 5)

= - ln (26 + 5) + 56 + \frac{1}{2} ln (26 + 5)

Simplify

- ln (26 + 5) + 56 + \frac{1}{2} ln (26 + 5) : - \frac{1}{2} ln (26 + 5) + 56

= - \frac{1}{2} ln (26 + 5) + 56

\int_{0}^{1} 6 cos (4 x) d x

\frac{d}{d x} (arcsin (e^{x}))

\int \frac{x ^{2}}{5 - x} d x

(2 z + e^{- z^{2}})^{6}

f (x) = 2 - x

What is the integral from 0 to 4 of sqrt(x(x+2)) ?
The integral from 0 to 4 of sqrt(x(x+2)) is -1/2 ln(2sqrt(6)+5)+5sqrt(6)