What is the integral of (x^3)/(sqrt(25-x^4)) ?

The integral of (x^3)/(sqrt(25-x^4)) is -1/2 sqrt(25-x^4)+C

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integral of (x^3)/(sqrt(25-x^4))

\int \frac{x ^{3}}{25 - x ^{4}} d x

- \frac{1}{2} 25 - x^{4} + C

Solution steps

\int \frac{x ^{3}}{25 - x ^{4}} d x

Apply u-substitution

= \int - \frac{1}{4 u} 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 \frac{1}{u} d u

Apply the Power Rule

= - \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 25 - x^{4}

= - \frac{1}{4} \cdot 2 (25 - x^{4})^{\frac{1}{2}}

Simplify

- \frac{1}{4} \cdot 2 (25 - x^{4})^{\frac{1}{2}} : - \frac{1}{2} 25 - x^{4}

= - \frac{1}{2} 25 - x^{4}

Add a constant to the solution

= - \frac{1}{2} 25 - x^{4} + C

What is the integral of (x^3)/(sqrt(25-x^4)) ?
The integral of (x^3)/(sqrt(25-x^4)) is -1/2 sqrt(25-x^4)+C