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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = x^{3} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{9}{x + 4})

\frac{d}{d x} (\frac{2 x + 5}{7 x ^{2}})

\frac{d}{d x} (lo g_{6} (cot (8 x)))

f (x) = sin (sin (x)), at x = π

x \to 0 lim (\frac{tan ( 3 x ) - sin ( 3 x )}{x ^{3}})

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