What is the integral of (3x)/(16-9x^2) ?

The integral of (3x)/(16-9x^2) is -1/6 ln|16-9x^2|+C

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integral of (3x)/(16-9x^2)

\int \frac{3 x}{16 - 9 x ^{2}} d x

- \frac{1}{6} ln 16 - 9 x^{2} + C

Solution steps

\int \frac{3 x}{16 - 9 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{16 - 9 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{18 u} d u

= 3 \cdot \int - \frac{1}{18 u} d u

Take the constant out:

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

= 3 (- \frac{1}{18} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{18} ln ∣ u ∣)

Substitute back

u = 16 - 9 x^{2}

= 3 (- \frac{1}{18} ln 16 - 9 x^{2})

Simplify

3 (- \frac{1}{18} ln 16 - 9 x^{2}) : - \frac{1}{6} ln 16 - 9 x^{2}

= - \frac{1}{6} ln 16 - 9 x^{2}

Add a constant to the solution

= - \frac{1}{6} ln 16 - 9 x^{2} + C

x \to 0 + lim (x ln (x^{4}))

\frac{\partial}{\partial x} (x ln (\frac{x ^{2}}{y ^{3}}))

\int_{- 2}^{3} \frac{26}{x ^{4}} d x

\frac{d}{d x} (\frac{1}{1 + x ^{4}})

What is the integral of (3x)/(16-9x^2) ?
The integral of (3x)/(16-9x^2) is -1/6 ln|16-9x^2|+C