What is the integral of x^3(1-x^4)^5 ?

The integral of x^3(1-x^4)^5 is -1/24 (1-x^4)^6+C

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integral of x^3(1-x^4)^5

\int x^{3} (1 - x^{4})^{5} d x

- \frac{1}{24} (1 - x^{4})^{6} + C

Solution steps

\int x^{3} (1 - x^{4})^{5} d x

Apply u-substitution

: \int - \frac{u ^{5}}{4} d u

= \int - \frac{u ^{5}}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

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

Substitute back

u = 1 - x^{4}

= - \frac{1}{4} \cdot \frac{( 1 - x ^{4} ) ^{6}}{6}

Simplify

- \frac{1}{4} \cdot \frac{( 1 - x ^{4} ) ^{6}}{6} : - \frac{1}{24} (1 - x^{4})^{6}

= - \frac{1}{24} (1 - x^{4})^{6}

Add a constant to the solution

= - \frac{1}{24} (1 - x^{4})^{6} + C

\int (\frac{7}{1 - x ^{2}}) d x

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

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

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

\frac{1}{4} q^{^{''}} + 20 q^{^{'}} + 300 q = 0

What is the integral of x^3(1-x^4)^5 ?
The integral of x^3(1-x^4)^5 is -1/24 (1-x^4)^6+C