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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u ^{2}} d u

= 3 \cdot \int \frac{1}{3 u ^{2}} 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 ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 3 \cdot \frac{1}{3} (- \frac{1}{u})

Substitute back

u = x^{3} + 1

= 3 \cdot \frac{1}{3} (- \frac{1}{x ^{3} + 1})

Simplify

3 \cdot \frac{1}{3} (- \frac{1}{x ^{3} + 1}) : - \frac{1}{x ^{3} + 1}

= - \frac{1}{x ^{3} + 1}

Add a constant to the solution

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

x \to 0 + lim (x^{4} ln (x))

2 cos (2 y)

\int_{0}^{2} 4 x^{2} d x

\frac{\partial}{\partial x} (ln (1 - 2 x))

\int_{1}^{2} \frac{1}{2 - x} d x

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