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

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

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

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

- \frac{1}{2 ( 1 + x ^{3} ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 3 \cdot \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = 1 + x^{3}

= 3 \cdot \frac{1}{3} (- \frac{1}{2 ( 1 + x ^{3} ) ^{2}})

Simplify

3 \cdot \frac{1}{3} (- \frac{1}{2 ( 1 + x ^{3} ) ^{2}}) : - \frac{1}{2 ( 1 + x ^{3} ) ^{2}}

= - \frac{1}{2 ( 1 + x ^{3} ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( 1 + x ^{3} ) ^{2}} + C

\frac{\partial}{\partial x} (e^{x} y^{2})

\frac{d x}{d t} = \frac{a}{x t}

\int - \frac{1}{2 x + 2} d x

x \to 2 - lim (\frac{6}{x - 2})

f (x) = 6 x^{4}

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