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

The integral from 0 to 2 of x^2sqrt(4x^2+1) is (-ln(4+sqrt(17))+132sqrt(17))/(64)

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

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

\frac{- ln ( 4 + 17 ) + 132 17}{64}

Decimal Notation

8.47117 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (4)} \frac{1}{8} sec^{3} (u) tan^{2} (u) d u

= \int_{0}^{a r c t a n (4)} \frac{1}{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{1}{8} \cdot \int_{0}^{a r c t a n (4)} sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (4)} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

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

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (4)} - 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{1}{8} (- \int_{0}^{a r c t a n (4)} sec^{3} (u) d u + \int_{0}^{a r c t a n (4)} sec^{5} (u) d u)

\int_{0}^{a r c t a n (4)} sec^{3} (u) d u = 217 + \frac{1}{2} ln (4 + 17)

\int_{0}^{a r c t a n (4)} sec^{5} (u) d u = 1717 + \frac{3}{4} (217 + \frac{1}{2} ln (4 + 17))

= \frac{1}{8} (- (217 + \frac{1}{2} ln (4 + 17)) + 1717 + \frac{3}{4} (217 + \frac{1}{2} ln (4 + 17)))

Simplify

\frac{1}{8} (- (217 + \frac{1}{2} ln (4 + 17)) + 1717 + \frac{3}{4} (217 + \frac{1}{2} ln (4 + 17))) : \frac{- ln ( 4 + 17 ) + 132 17}{64}

= \frac{- ln ( 4 + 17 ) + 132 17}{64}

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What is the integral from 0 to 2 of x^2sqrt(4x^2+1) ?
The integral from 0 to 2 of x^2sqrt(4x^2+1) is (-ln(4+sqrt(17))+132sqrt(17))/(64)