What is the integral of (3x^2)/(5x^3-1) ?

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

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integral of (3x^2)/(5x^3-1)

\int \frac{3 x ^{2}}{5 x ^{3} - 1} d x

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

Solution steps

\int \frac{3 x ^{2}}{5 x ^{3} - 1} 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 ^{2}}{5 x ^{3} - 1} d x

Apply u-substitution

: \int \frac{1}{15 u} d u

= 3 \cdot \int \frac{1}{15 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{15} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{15} ln ∣ u ∣

Substitute back

u = 5 x^{3} - 1

= 3 \cdot \frac{1}{15} ln 5 x^{3} - 1

Simplify

3 \cdot \frac{1}{15} ln 5 x^{3} - 1 : \frac{1}{5} ln 5 x^{3} - 1

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

Add a constant to the solution

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

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

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

\frac{\partial}{\partial x} (e^{x y} x)

\frac{d}{d x} (sin (x + \frac{π}{2}))

\int (x^{6} - 3 x^{2})^{4} (x^{5} - x) d x

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