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

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

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

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

ln x^{3} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 1

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

Simplify

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

= ln x^{3} + 1

Add a constant to the solution

= ln x^{3} + 1 + C

y (x - b) + y^{^{'}} b x = 0

y^{^{'}} = \frac{- y ^{3} + 14 x - 8}{3 x y ^{2}}

\frac{d}{d x} (\frac{2}{2 x ^{4} - 5})

\frac{d}{d x} (\frac{( x + 1 ) ^{2} + ( x - 1 ) ^{2}}{x ^{2}})

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

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