What is the integral from 0 to 1 of x^2sqrt(x^2+3) ?

The integral from 0 to 1 of x^2sqrt(x^2+3) is 9(5/36-1/16 ln(3))

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integral from 0 to 1 of x^2sqrt(x^2+3)

\int_{0}^{1} x^{2} x^{2} + 3 d x

9 (\frac{5}{36} - \frac{1}{16} ln (3))

Decimal

0.63203 \dots

Solution steps

\int_{0}^{1} x^{2} x^{2} + 3 d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 9 sec^{3} (u) tan^{2} (u) d u

= \int_{0}^{\frac{π}{6}} 9 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

= 9 \cdot \int_{0}^{\frac{π}{6}} sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int_{0}^{\frac{π}{6}} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

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

= 9 \cdot \int_{0}^{\frac{π}{6}} - 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

= 9 (- \int_{0}^{\frac{π}{6}} sec^{3} (u) d u + \int_{0}^{\frac{π}{6}} sec^{5} (u) d u)

\int_{0}^{\frac{π}{6}} sec^{3} (u) d u = \frac{1}{3} + \frac{1}{4} ln (3)

\int_{0}^{\frac{π}{6}} sec^{5} (u) d u = \frac{17}{36} + \frac{3}{16} ln (3)

= 9 (- (\frac{1}{3} + \frac{1}{4} ln (3)) + \frac{17}{36} + \frac{3}{16} ln (3))

Expand

- (\frac{1}{3} + \frac{1}{4} ln (3)) + \frac{17}{36} + \frac{3}{16} ln (3) : \frac{5}{36} - \frac{1}{16} ln (3)

= 9 (\frac{5}{36} - \frac{1}{16} ln (3))

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What is the integral from 0 to 1 of x^2sqrt(x^2+3) ?
The integral from 0 to 1 of x^2sqrt(x^2+3) is 9(5/36-1/16 ln(3))