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

The integral from 0 to 4 of x/(sqrt(1+x^2)) is sqrt(17)-1

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integral from 0 to 4 of x/(sqrt(1+x^2))

\int_{0}^{4} \frac{x}{1 + x ^{2}} d x

17 - 1

Decimal

3.12310 \dots

Solution steps

\int_{0}^{4} \frac{x}{1 + x ^{2}} d x

Apply u-substitution

: \int_{1}^{17} \frac{1}{2 u} d u

= \int_{1}^{17} \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_{1}^{17} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{17}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{17}

Apply radical rule:

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

= \frac{1}{2} [2 u]_{1}^{17}

Compute the boundaries:

217 - 2

= \frac{1}{2} (217 - 2)

Simplify

= 17 - 1

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What is the integral from 0 to 4 of x/(sqrt(1+x^2)) ?
The integral from 0 to 4 of x/(sqrt(1+x^2)) is sqrt(17)-1