What is the integral of x/(3x^2-5) ?

The integral of x/(3x^2-5) is 1/6 ln|3x^2-5|+C

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integral of x/(3x^2-5)

\int \frac{x}{3 x ^{2} - 5} d x

\frac{1}{6} ln 3 x^{2} - 5 + C

Solution steps

\int \frac{x}{3 x ^{2} - 5} 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} - 5

= \frac{1}{6} ln 3 x^{2} - 5

Add a constant to the solution

= \frac{1}{6} ln 3 x^{2} - 5 + C

\frac{d y}{d x} = 3 x y^{2}

(x - y^{4} + y^{2} sin (x)) d x = (4 x y^{3} + 2 y cos (x)) d y

y^{^{''}} + 4 y = 7 sin (2 x)

5 x y (\frac{d y}{d x}) = 2

\frac{3}{t ^{4}}

What is the integral of x/(3x^2-5) ?
The integral of x/(3x^2-5) is 1/6 ln|3x^2-5|+C