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

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

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

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

\frac{- 4 ln ( \frac{3 + 5}{2} ) + 3 5}{2}

Decimal Notation

1.42925 \dots

Solution steps

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

Apply Trigonometric Substitution

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

= \int_{- a r c t a n (\frac{1}{2})}^{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

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

Rewrite using trig identities

= 16 \cdot \int_{- a r c t a n (\frac{1}{2})}^{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)

= 16 \cdot \int_{- a r c t a n (\frac{1}{2})}^{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

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

\int_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} sec^{3} (u) d u = \frac{5}{4} + \frac{1}{2} ln (\frac{3 + 5}{2})

\int_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} sec^{5} (u) d u = \frac{11 5 + 12 ln ( \frac{3 + 5}{2} )}{32}

= 16 - (\frac{5}{4} + \frac{1}{2} ln (\frac{3 + 5}{2})) + \frac{11 5 + 12 ln ( \frac{3 + 5}{2} )}{32}

Simplify

16 - (\frac{5}{4} + \frac{1}{2} ln (\frac{3 + 5}{2})) + \frac{11 5 + 12 ln ( \frac{3 + 5}{2} )}{32} : \frac{- 4 ln ( \frac{3 + 5}{2} ) + 3 5}{2}

= \frac{- 4 ln ( \frac{3 + 5}{2} ) + 3 5}{2}

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