What is the integral from 0 to 5 of t/(sqrt(11+t^2)) ?

The integral from 0 to 5 of t/(sqrt(11+t^2)) is 6-sqrt(11)

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integral from 0 to 5 of t/(sqrt(11+t^2))

\int_{0}^{5} \frac{t}{11 + t ^{2}} d t

6 - 11

Decimal

2.68337 \dots

Solution steps

\int_{0}^{5} \frac{t}{11 + t ^{2}} d t

Apply u-substitution

: \int_{11}^{36} \frac{1}{2 u} d u

= \int_{11}^{36} \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_{11}^{36} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{11}^{36}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{11}^{36}

Apply radical rule:

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

= \frac{1}{2} [2 u]_{11}^{36}

Compute the boundaries:

12 - 211

= \frac{1}{2} (12 - 211)

Simplify

= 6 - 11

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What is the integral from 0 to 5 of t/(sqrt(11+t^2)) ?
The integral from 0 to 5 of t/(sqrt(11+t^2)) is 6-sqrt(11)