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

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

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

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

2 - 3

Decimal

0.26794 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{4} \frac{1}{2 u} d u

= \int_{3}^{4} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{3}^{4} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{3}^{4}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{3}^{4}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{3}^{4}

Compute the boundaries:

4 - 23

= \frac{1}{2} (4 - 23)

Simplify

= 2 - 3

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