What is the integral from 0 to 2 of t/(sqrt(5+t^2)) ?

The integral from 0 to 2 of t/(sqrt(5+t^2)) is 3-sqrt(5)

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integral from 0 to 2 of t/(sqrt(5+t^2))

\int_{0}^{2} \frac{t}{5 + t ^{2}} d t

3 - 5

Solution steps

\int_{0}^{2} \frac{t}{5 + t ^{2}} d t

Apply u-substitution

: \int_{5}^{9} \frac{1}{2 u} d u

= \int_{5}^{9} \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_{5}^{9} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{5}^{9}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{5}^{9}

Apply radical rule:

= \frac{1}{2} [2 u]_{5}^{9}

Compute the boundaries:

6 - 25

= \frac{1}{2} (6 - 25)

Simplify

= 3 - 5

\int_{0}^{\frac{π}{4}} sin (2 x + \frac{π}{2}) d x

\int_{0}^{\frac{π}{2}} x^{2} \cdot cos (x) d x

\int_{0}^{1} x^{2} \cdot 1 - x^{2} d x

\int_{0}^{5} (4 - x^{2}) d x

What is the integral from 0 to 2 of t/(sqrt(5+t^2)) ?
The integral from 0 to 2 of t/(sqrt(5+t^2)) is 3-sqrt(5)