What is the integral from 0 to 1 of x^2sqrt(9+4x^2) ?

The integral from 0 to 1 of x^2sqrt(9+4x^2) is (-81ln((2+sqrt(13))/3)+34sqrt(13))/(64)

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integral from 0 to 1 of x^2sqrt(9+4x^2)

\int_{0}^{1} x^{2} 9 + 4 x^{2} d x

\frac{- 81 ln ( \frac{2 + 13}{3} ) + 34 13}{64}

Decimal

1.12424 \dots

Solution steps

\int_{0}^{1} x^{2} 9 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{2}{3})} \frac{81}{8} sec^{3} (u) tan^{2} (u) d u

= \int_{0}^{a r c t a n (\frac{2}{3})} \frac{81}{8} 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

= \frac{81}{8} \cdot \int_{0}^{a r c t a n (\frac{2}{3})} sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{81}{8} \cdot \int_{0}^{a r c t a n (\frac{2}{3})} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

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

= \frac{81}{8} \cdot \int_{0}^{a r c t a n (\frac{2}{3})} - 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

= \frac{81}{8} (- \int_{0}^{a r c t a n (\frac{2}{3})} sec^{3} (u) d u + \int_{0}^{a r c t a n (\frac{2}{3})} sec^{5} (u) d u)

\int_{0}^{a r c t a n (\frac{2}{3})} sec^{3} (u) d u = \frac{13}{9} + \frac{1}{2} ln (\frac{2 + 13}{3})

\int_{0}^{a r c t a n (\frac{2}{3})} sec^{5} (u) d u = \frac{106 13 + 243 ln ( \frac{2 + 13}{3} )}{648}

= \frac{81}{8} - (\frac{13}{9} + \frac{1}{2} ln (\frac{2 + 13}{3})) + \frac{106 13 + 243 ln ( \frac{2 + 13}{3} )}{648}

Simplify

\frac{81}{8} - (\frac{13}{9} + \frac{1}{2} ln (\frac{2 + 13}{3})) + \frac{106 13 + 243 ln ( \frac{2 + 13}{3} )}{648} : \frac{- 81 ln ( \frac{2 + 13}{3} ) + 34 13}{64}

= \frac{- 81 ln ( \frac{2 + 13}{3} ) + 34 13}{64}

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What is the integral from 0 to 1 of x^2sqrt(9+4x^2) ?
The integral from 0 to 1 of x^2sqrt(9+4x^2) is (-81ln((2+sqrt(13))/3)+34sqrt(13))/(64)