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

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

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integral of (3x^2)/(2x^3+1)

\int \frac{3 x ^{2}}{2 x ^{3} + 1} d x

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

Solution steps

\int \frac{3 x ^{2}}{2 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}}{2 x ^{3} + 1} d x

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 2 x^{3} + 1

= 3 \cdot \frac{1}{6} ln 2 x^{3} + 1

Simplify

3 \cdot \frac{1}{6} ln 2 x^{3} + 1 : \frac{1}{2} ln 2 x^{3} + 1

= \frac{1}{2} ln 2 x^{3} + 1

Add a constant to the solution

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

2 x^{5} - 11 x^{3}

\frac{\partial}{\partial x} (\frac{A ( x , v , t )}{( x - v t ) ^{2} + 1})

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

\frac{1}{7 s ^{2} + 37.4 s + 19980}

n = 0 \sum \infty (- 0.5)^{n}

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