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

The integral from 0 to 4 of x/(sqrt(x^2+4)) is 2(sqrt(5)-1)

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

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

2 (5 - 1)

Decimal

2.47213 \dots

Solution steps

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

Apply u-substitution

= \int_{4}^{20} \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_{4}^{20} \frac{1}{u} d u

Apply the Power Rule

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{4}^{20}

Apply radical rule:

= \frac{1}{2} [2 u]_{4}^{20}

Compute the boundaries:

45 - 4

= \frac{1}{2} (45 - 4)

Simplify

= 2 (5 - 1)

What is the integral from 0 to 4 of x/(sqrt(x^2+4)) ?
The integral from 0 to 4 of x/(sqrt(x^2+4)) is 2(sqrt(5)-1)