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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

: \int - \frac{1}{2 u} d u

= 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

: 2 u^{\frac{1}{2}}

= 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

\frac{d}{d x} ((x + 3)^{3})

\int - 3 x cos (4 x) d x

\int (- 5 x^{4} + 8 x^{3}) (- 2 x^{5} + 4 x^{4})^{6} d x

f (x) = 3 x^{2} + 6

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

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