What is the integral of 5 x/(sqrt(16-x^2)) ?

The integral of 5 x/(sqrt(16-x^2)) is -5sqrt(16-x^2)+C

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integral of 5 x/(sqrt(16-x^2))

\int 5 \frac{x}{16 - x ^{2}} d x

- 5 16 - x^{2} + C

Solution steps

\int 5 \cdot \frac{x}{16 - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{x}{16 - x ^{2}} d x

Apply u-substitution

= 5 \cdot \int - \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

= 5 (- \frac{1}{2} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 16 - x^{2}

= 5 (- \frac{1}{2} \cdot 2 (16 - x^{2})^{\frac{1}{2}})

Simplify

5 (- \frac{1}{2} \cdot 2 (16 - x^{2})^{\frac{1}{2}}) : - 5 16 - x^{2}

= - 5 16 - x^{2}

Add a constant to the solution

= - 5 16 - x^{2} + C

What is the integral of 5 x/(sqrt(16-x^2)) ?
The integral of 5 x/(sqrt(16-x^2)) is -5sqrt(16-x^2)+C