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

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

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

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

2 - 1

Decimal Notation

0.41421 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Apply radical rule:

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

Compute the boundaries:

22 - 2

= \frac{1}{2} (22 - 2)

Simplify

= 2 - 1

\frac{d}{d x} (1 + 2 sin (\frac{x ^{2}}{2}))

\int \frac{4}{x ^{2} + 16} d x

\int \frac{2 ( x + 2 )}{5} d x

f (x) = \frac{x ^{3} + 2}{3}, at x = - 3

\frac{d}{d x} (x^{4} + y^{4})

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