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

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

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

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

\frac{1}{6} ln 3 x^{2} - 16 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 x^{2} - 16

= \frac{1}{6} ln 3 x^{2} - 16

Add a constant to the solution

= \frac{1}{6} ln 3 x^{2} - 16 + C

\int 2 cot (x) d x

\frac{d}{d x} (\frac{x y}{( x ^{2} - y ^{2} ) ^{2}})

x \to - 2 + lim (\frac{1}{x ^{2} - 4})

f (x) = x + 8, (1, 3)

n = 1 \sum \infty 3^{n} 5^{1 - n}

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