Evaluate the integral: I = integral (x^5)/(sqrt(1-x^4)) dx

Evaluate the integral: $$ I = \int \frac{x^5}{\sqrt{1-x^4}}\,dx. $$

Step 1 · Substitution

Let $t=x^2$, so $dt=2x\,dx$. Writing $x^5\,dx=x^4\cdot(x\,dx)=t^2\cdot\dfrac{dt}2$, and $\sqrt{1-x^4}=\sqrt{1-t^2}$:

$$ I = \int \frac{t^2}{\sqrt{1-t^2}}\cdot\frac{dt}{2} = \frac12\int \frac{t^2}{\sqrt{1-t^2}}\,dt. $$

Step 2 · Standard result

Using the known antiderivative $\displaystyle\int \frac{t^2}{\sqrt{1-t^2}}\,dt = -\frac{t}{2}\sqrt{1-t^2}+\frac12\arcsin t$:

$$ I = \frac12\left(-\frac{t}{2}\sqrt{1-t^2}+\frac12\arcsin t\right) + C = -\frac{t}{4}\sqrt{1-t^2} + \frac14\arcsin t + C. $$

Step 3 · Substitute back

With $t=x^2$:

$$ I = -\frac{x^2}{4}\sqrt{1-x^4} + \frac14\arcsin\left(x^2\right) + C. $$

Summary Table

Additional Notes

• The domain requires $-1\leq x\leq1$, where $\sqrt{1-x^4}$ is real.
• The standard antiderivative $\int t^2/\sqrt{1-t^2}\,dt=-\tfrac t2\sqrt{1-t^2}+\tfrac12\arcsin t$ can itself be derived by integration by parts with $u=t$, $dv=t/\sqrt{1-t^2}\,dt$, or by the substitution $t=\sin\phi$ followed by a power-reduction identity, as in Problem 181.
• The result can be checked by differentiating $-\tfrac{x^2}{4}\sqrt{1-x^4}+\tfrac14\arcsin(x^2)$: the product-rule derivative of the first term is $-\tfrac{x}{2}\sqrt{1-x^4}+\tfrac{x^5}{2\sqrt{1-x^4}}$, and the derivative of the second term is $\tfrac{x}{2\sqrt{1-x^4}}$; combining over the common denominator $\sqrt{1-x^4}$ gives numerator $-x(1-x^4)+x^5+x=x^5$, so the whole expression reduces exactly to $x^5/\sqrt{1-x^4}$.