Evaluate the integral: I = integral

Evaluate the integral: $$ I = \int \frac{x+x^3}{\sqrt{1+x^2-x^4}}\,dx. $$

Step 1 · Substitute $u=x^2$

Since $x+x^3=x(1+x^2)$, let $u=x^2$, $du=2x\,dx$:

$$ I = \frac12\int \frac{1+u}{\sqrt{1+u-u^2}}\,du. $$

Step 2 · Split the numerator

Since $\dfrac{d}{du}(1+u-u^2)=1-2u$, write $1+u=-\tfrac12(1-2u)+\tfrac32$:

$$ I = \frac12\left[-\frac12\int \frac{1-2u}{\sqrt{1+u-u^2}}\,du + \frac32\int \frac{du}{\sqrt{1+u-u^2}}\right]. $$

Step 3 · Integrate each piece

The first piece is a direct square-root match: $\int(1-2u)/\sqrt{1+u-u^2}\,du=2\sqrt{1+u-u^2}$, so $-\tfrac12$ of it is $-\sqrt{1+u-u^2}$. For the second, complete the square:

$$ 1+u-u^2 = \frac54 - \left(u-\frac12\right)^2, \qquad \int \frac{du}{\sqrt{1+u-u^2}} = \arcsin\left(\frac{2u-1}{\sqrt5}\right). $$

So the bracket equals $-\sqrt{1+u-u^2}+\tfrac32\arcsin\!\left(\tfrac{2u-1}{\sqrt5}\right)$.

Step 4 · Combine and substitute back

$$ I = \frac12\left[-\sqrt{1+u-u^2}+\frac32\arcsin\left(\frac{2u-1}{\sqrt5}\right)\right] + C = -\frac12\sqrt{1+u-u^2}+\frac34\arcsin\left(\frac{2u-1}{\sqrt5}\right) + C. $$

With $u=x^2$ and $1+u-u^2=1+x^2-x^4$:

$$ I = -\frac12\sqrt{1+x^2-x^4} + \frac34\arcsin\left(\frac{2x^2-1}{\sqrt5}\right) + C. $$

Summary Table

Additional Notes

• Correction: the source tried several substitutions (all abandoned as "not simplifying") before finally reaching the correct $u=x^2$ route in its last attempt — but even that attempt trailed off after setting up the trigonometric substitution, never combining the pieces into a final answer. The setup shown (splitting $1+u$, completing the square, and the $\theta$-substitution) was on the right track; it just needed to be carried through to completion, as done above.
• The domain requires $1+x^2-x^4\geq0$, i.e. $x^2\leq\tfrac{1+\sqrt5}{2}$ (the golden ratio), so $|x|\leq\sqrt{\tfrac{1+\sqrt5}{2}}$.
• The result was verified directly by differentiation: writing $s=\sqrt{1+x^2-x^4}$, the derivative of $-\tfrac12s$ is $(2x^3-x)/(2s)$, and the derivative of $\tfrac34\arcsin\!\left(\tfrac{2x^2-1}{\sqrt5}\right)$ (using $5-(2x^2-1)^2=4(1+x^2-x^4)$) is $3x/(2s)$; summing gives $(2x^3-x+3x)/(2s)=(2x^3+2x)/(2s)=(x^3+x)/s$, matching the integrand.