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

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

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

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

5 - 1

Decimal Notation

1.23606 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{5} \frac{1}{2 u} d u

= \int_{1}^{5} \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}^{5} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{5}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{5}

Apply radical rule:

= \frac{1}{2} [2 u]_{1}^{5}

Compute the boundaries:

25 - 2

= \frac{1}{2} (25 - 2)

Simplify

= 5 - 1

\frac{d}{d x} f (x) = 3 x^{2} - 5 x + 2

\frac{d y}{d x}, x^{5} y + 5 x y^{5} = x + y

x \to 2 + lim (cos (x - 2))

\int y e^{y^{2}} d y

x y^{^{'}} - 2 y = x^{2}

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