What is the integral from 0 to 4 of p/(sqrt(9+p^2)) ?

The integral from 0 to 4 of p/(sqrt(9+p^2)) is 2

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integral from 0 to 4 of p/(sqrt(9+p^2))

\int_{0}^{4} \frac{p}{9 + p ^{2}} d p

2

Solution steps

\int_{0}^{4} \frac{p}{9 + p ^{2}} d p

Apply u-substitution

: \int_{9}^{25} \frac{1}{2 u} d u

= \int_{9}^{25} \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_{9}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{25}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{25}

Apply radical rule:

= \frac{1}{2} [2 u]_{9}^{25}

Compute the boundaries:

4

= \frac{1}{2} \cdot 4

Simplify

= 2

\int_{0}^{4} 2 x^{2} + 5 d x

\int_{- 1}^{1} ∣ x + 2 ∣ d x

\int_{- 2021}^{2021} x^{5} d x

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

What is the integral from 0 to 4 of p/(sqrt(9+p^2)) ?
The integral from 0 to 4 of p/(sqrt(9+p^2)) is 2