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

The integral from 0 to 1 of x^2sqrt(1+x^2) is (-sqrt(2)ln(1+sqrt(2))+6)/(8sqrt(2))

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

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

\frac{- 2 ln ( 1 + 2 ) + 6}{8 2}

Decimal

0.42015 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} sec^{3} (u) tan^{2} (u) d u

= \int_{0}^{\frac{π}{4}} sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

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

= \int_{0}^{\frac{π}{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

= - \int_{0}^{\frac{π}{4}} sec^{3} (u) d u + \int_{0}^{\frac{π}{4}} sec^{5} (u) d u

\int_{0}^{\frac{π}{4}} sec^{3} (u) d u = \frac{2 + ln ( 1 + 2 )}{2}

\int_{0}^{\frac{π}{4}} sec^{5} (u) d u = \frac{3 2 ln ( 1 + 2 ) + 14}{8 2}

= - \frac{2 + ln ( 1 + 2 )}{2} + \frac{3 2 ln ( 1 + 2 ) + 14}{8 2}

Simplify

- \frac{2 + ln ( 1 + 2 )}{2} + \frac{3 2 ln ( 1 + 2 ) + 14}{8 2} : \frac{- 2 ln ( 1 + 2 ) + 6}{8 2}

= \frac{- 2 ln ( 1 + 2 ) + 6}{8 2}

\int_{- π}^{π} x^{4} sin (x) d x

\int_{0}^{\frac{1}{2}} sin (π x) d x

\int_{4}^{9} \frac{x}{x - 1} d x

\int_{\frac{1}{2}}^{\frac{1}{3}} cos (π x) d x

\int_{0}^{π} x cos (x^{2}) d x

What is the integral from 0 to 1 of x^2sqrt(1+x^2) ?
The integral from 0 to 1 of x^2sqrt(1+x^2) is (-sqrt(2)ln(1+sqrt(2))+6)/(8sqrt(2))