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

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

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

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

2 - 1

Decimal Notation

0.41421 \dots

Solution steps

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

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\int_{- 1}^{1} (6 - 6 x^{2})^{2} d x

x \to \infty lim (arctan (5) - arctan (x + 1))

8 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

\frac{d y}{d x} + y^{3} x + 2 y = 0

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