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

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

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

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

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

Decimal

1.07357 \dots

Solution steps

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

Complete the square

2 x + x^{2} : (x + 1)^{2} - 1

= \int_{0}^{1} (x + 1)^{2} - 1 d x

Apply u-substitution

: \int_{1}^{2} u^{2} - 1 d u

= \int_{1}^{2} u^{2} - 1 d u

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

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

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

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}{3}} sec (v) d v + \int_{0}^{\frac{π}{3}} sec^{3} (v) d v

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

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

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

Simplify

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

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

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