What is the integral from 3 to 7 of t/(sqrt(7+t^2)) ?

The integral from 3 to 7 of t/(sqrt(7+t^2)) is 2(sqrt(14)-2)

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integral from 3 to 7 of t/(sqrt(7+t^2))

\int_{3}^{7} \frac{t}{7 + t ^{2}} d t

2 (14 - 2)

Decimal Notation

3.48331 \dots

Solution steps

\int_{3}^{7} \frac{t}{7 + t ^{2}} d t

Apply u-substitution

: \int_{16}^{56} \frac{1}{2 u} d u

= \int_{16}^{56} \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_{16}^{56} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{16}^{56}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{16}^{56}

Apply radical rule:

= \frac{1}{2} [2 u]_{16}^{56}

Compute the boundaries:

414 - 8

= \frac{1}{2} (414 - 8)

Simplify

= 2 (14 - 2)

\int \frac{3 - x}{2 x ^{2} + x - 1} d x

x v^{'} + v = \frac{2 v}{1 - v ^{2}}

f (x) = 2 x + 77, at x = 2

x \to 0 - lim (\frac{( 6 x )}{x ^{4}})

y^{^{''}} - 8 y^{^{'}} + 7 y = 0

What is the integral from 3 to 7 of t/(sqrt(7+t^2)) ?
The integral from 3 to 7 of t/(sqrt(7+t^2)) is 2(sqrt(14)-2)