What is the integral of (5x)/(16-4x^2) ?

The integral of (5x)/(16-4x^2) is -5/8 ln|16-4x^2|+C

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integral of (5x)/(16-4x^2)

\int \frac{5 x}{16 - 4 x ^{2}} d x

- \frac{5}{8} ln 16 - 4 x^{2} + C

Solution steps

\int \frac{5 x}{16 - 4 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 - 4 x ^{2}} d x

Apply u-substitution

= 5 \cdot \int - \frac{1}{8 u} d u

Take the constant out:

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

= 5 (- \frac{1}{8} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 (- \frac{1}{8} ln ∣ u ∣)

Substitute back

u = 16 - 4 x^{2}

= 5 (- \frac{1}{8} ln 16 - 4 x^{2})

Simplify

5 (- \frac{1}{8} ln 16 - 4 x^{2}) : - \frac{5}{8} ln 16 - 4 x^{2}

= - \frac{5}{8} ln 16 - 4 x^{2}

Add a constant to the solution

= - \frac{5}{8} ln 16 - 4 x^{2} + C

\int_{12}^{30} \frac{1}{30} d x

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\frac{d}{d x} (4 cos^{5} (x))

\frac{\partial}{\partial y} (2 e^{x} cos (yz))

\frac{1}{3} (\frac{d x}{d t}) - 2 e^{- t} = - x

What is the integral of (5x)/(16-4x^2) ?
The integral of (5x)/(16-4x^2) is -5/8 ln|16-4x^2|+C