What is the integral of ((24x)/(3-4x^2)) ?

The integral of ((24x)/(3-4x^2)) is -3ln|3-4x^2|+C

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integral of ((24x)/(3-4x^2))

\int (\frac{24 x}{3 - 4 x ^{2}}) d x

- 3 ln 3 - 4 x^{2} + C

Solution steps

\int \frac{24 x}{3 - 4 x ^{2}} d x

Take the constant out:

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

= 24 \cdot \int \frac{x}{3 - 4 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{8 u} d u

= 24 \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

= 24 (- \frac{1}{8} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 24 (- \frac{1}{8} ln ∣ u ∣)

Substitute back

u = 3 - 4 x^{2}

= 24 (- \frac{1}{8} ln 3 - 4 x^{2})

Simplify

24 (- \frac{1}{8} ln 3 - 4 x^{2}) : - 3 ln 3 - 4 x^{2}

= - 3 ln 3 - 4 x^{2}

Add a constant to the solution

= - 3 ln 3 - 4 x^{2} + C

\frac{d}{d x} (2 (cos (\frac{x}{2}) - \frac{2}{2}))

\int_{2}^{3} 4 e^{4 x} d x

y^{^{'}} + 3 y = t e^{4 t}

\int_{16}^{25} \frac{x - 16}{x} d x

What is the integral of ((24x)/(3-4x^2)) ?
The integral of ((24x)/(3-4x^2)) is -3ln|3-4x^2|+C