What is the integral from 0 to 1 of 2x^2sqrt(x^2+4) ?

The integral from 0 to 1 of 2x^2sqrt(x^2+4) is (-8ln((1+sqrt(5))/2)+3sqrt(5))/(2)

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integral from 0 to 1 of 2x^2sqrt(x^2+4)

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

\frac{- 8 ln ( \frac{1 + 5}{2} ) + 3 5}{2}

Decimal Notation

1.42925 \dots

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{1}{2})} 16 sec^{3} (u) tan^{2} (u) d u

= 2 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} 16 sec^{3} (u) tan^{2} (u) d u

Take the constant out:

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

= 2 \cdot 16 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= 2 \cdot 16 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 2 \cdot 16 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} - sec^{3} (u) + sec^{5} (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 \cdot 16 (- \int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) d u + \int_{0}^{a r c t a n (\frac{1}{2})} sec^{5} (u) d u)

\int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) d u = \frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2})

\int_{0}^{a r c t a n (\frac{1}{2})} sec^{5} (u) d u = \frac{11 5 + 24 ln ( \frac{1 + 5}{2} )}{64}

= 2 \cdot 16 - (\frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2})) + \frac{11 5 + 24 ln ( \frac{1 + 5}{2} )}{64}

Simplify

2 \cdot 16 - (\frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2})) + \frac{11 5 + 24 ln ( \frac{1 + 5}{2} )}{64} : \frac{- 8 ln ( \frac{1 + 5}{2} ) + 3 5}{2}

= \frac{- 8 ln ( \frac{1 + 5}{2} ) + 3 5}{2}

\int e^{- 0.02 x} d x

\int \frac{1}{( 2 x - 1 )} d x

t^{2} \frac{d y}{d t} - t = 1 + y + y t

y^{^{''}} = ey

\frac{d}{d x} (e^{x + 5})

What is the integral from 0 to 1 of 2x^2sqrt(x^2+4) ?
The integral from 0 to 1 of 2x^2sqrt(x^2+4) is (-8ln((1+sqrt(5))/2)+3sqrt(5))/(2)