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

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

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

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

b^{2} + 1 - 1

Solution steps

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

Apply u-substitution

: \int_{1}^{b^{2} + 1} \frac{1}{2 u} d u

= \int_{1}^{b^{2} + 1} \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}^{b^{2} + 1} \frac{1}{u} d u

Apply the Power Rule

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

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

Apply radical rule:

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

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

Compute the boundaries:

2 b^{2} + 1 - 2

= \frac{1}{2} (2 b^{2} + 1 - 2)

Simplify

= b^{2} + 1 - 1

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