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

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

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

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

5 - 2

Decimal Notation

0.82185 \dots

Solution steps

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

Apply u-substitution

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

= \int_{2}^{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_{2}^{5} \frac{1}{u} d u

Apply the Power Rule

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

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

Apply radical rule:

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

Compute the boundaries:

25 - 22

= \frac{1}{2} (25 - 22)

Simplify

= 5 - 2

(x - \frac{2}{x})^{3}

\int_{0}^{7} e^{x} - 7 d x

\frac{\partial}{\partial x} (z^{2} (sin (2 x y)))

y^{^{'}} = y (x y^{3} + 3)

\int_{0}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

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