What is the integral of x(1-3x^2)^4 ?

The integral of x(1-3x^2)^4 is -1/30 (1-3x^2)^5+C

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integral of x(1-3x^2)^4

\int x (1 - 3 x^{2})^{4} d x

- \frac{1}{30} (1 - 3 x^{2})^{5} + C

Solution steps

\int x (1 - 3 x^{2})^{4} d x

Apply u-substitution

: \int - \frac{u ^{4}}{6} d u

= \int - \frac{u ^{4}}{6} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= - \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 1 - 3 x^{2}

= - \frac{1}{6} \cdot \frac{( 1 - 3 x ^{2} ) ^{5}}{5}

Simplify

- \frac{1}{6} \cdot \frac{( 1 - 3 x ^{2} ) ^{5}}{5} : - \frac{1}{30} (1 - 3 x^{2})^{5}

= - \frac{1}{30} (1 - 3 x^{2})^{5}

Add a constant to the solution

= - \frac{1}{30} (1 - 3 x^{2})^{5} + C

ln (x) \cdot x

\frac{\partial}{\partial u} (4 sin (u) cos (v))

x \to 0 + lim (tan (2 x))

(8 x + 9)^{- 2}

What is the integral of x(1-3x^2)^4 ?
The integral of x(1-3x^2)^4 is -1/30 (1-3x^2)^5+C