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

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

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integral from 0 to 3 of t/(sqrt(3+t^2))

\int_{0}^{3} \frac{t}{3 + t ^{2}} d t

3

Decimal

1.73205 \dots

Solution steps

\int_{0}^{3} \frac{t}{3 + t ^{2}} d t

Apply u-substitution

: \int_{3}^{12} \frac{1}{2 u} d u

= \int_{3}^{12} \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_{3}^{12} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{3}^{12}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{3}^{12}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{3}^{12}

Compute the boundaries:

23

= \frac{1}{2} \cdot 23

Simplify

= 3

f (x) = 4^{x}, at x = 0

\frac{d}{d x} (- 3 cos (x) + 4 sin (x))

y^{^{'}} - \frac{y}{x} = - x e^{- x}

f (x) = ln (\frac{3}{2}) x^{8}

\frac{d y}{d x} = 8 x y cos (x^{2})

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