Evaluate the integral: I = integral (x

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

Step 1 · Substitute $t=x^2$

Let $t=x^2$, $dt=2x\,dx$:

$$ I = \frac12\int \frac{dt}{\sqrt{1-3t-2t^2}}. $$

Step 2 · Complete the square

$$ 1-3t-2t^2 = -2\left(t^2+\frac32t\right)+1 = -2\left[\left(t+\frac34\right)^2-\frac{17}{16}\right]+1 = \frac{17}{8} - 2\left(t+\frac34\right)^2. $$

So:

$$ \sqrt{1-3t-2t^2} = \sqrt2\,\sqrt{\frac{17}{16}-\left(t+\frac34\right)^2}. $$

Step 3 · Apply the standard arcsine formula

Let $u=t+\tfrac34$, $a=\dfrac{\sqrt{17}}{4}$:

$$ I = \frac{1}{2\sqrt2}\int \frac{du}{\sqrt{a^2-u^2}} = \frac{1}{2\sqrt2}\arcsin\frac ua + C = \frac{1}{2\sqrt2}\arcsin\left(\frac{4u}{\sqrt{17}}\right) + C. $$

Step 4 · Substitute back

With $u=t+\tfrac34$, so $4u=4t+3$, and $t=x^2$:

$$ I = \frac{1}{2\sqrt2}\arcsin\left(\frac{4x^2+3}{\sqrt{17}}\right) + C. $$

Summary Table

Additional Notes

• Correction: the source claimed $2t^2+3t-1=(2t-1)(t+1)$, but expanding $(2t-1)(t+1)=2t^2+t-1$, not $2t^2+3t-1$ — the middle term is wrong. In fact the discriminant of $2t^2+3t-1$ is $9+8=17$, which is not a perfect square, so this quadratic has irrational roots and does not factor over the rationals at all (the same type of error as in Problem 192). Completing the square, as done above, is the correct route, and it produces a $\sqrt{17}$ in the final answer rather than the source's rational-looking (but invalid) factorization.
• The domain requires $1-3x^2-2x^4\geq0$, i.e. $x^2\leq\tfrac{-3+\sqrt{17}}{4}$, so $|x|\leq\sqrt{\tfrac{\sqrt{17}-3}{4}}$.
• The result was verified both symbolically and numerically: differentiating $\tfrac{1}{2\sqrt2}\arcsin\!\left(\tfrac{4x^2+3}{\sqrt{17}}\right)$ gives $\tfrac{2\sqrt2\,x}{\sqrt{17-(4x^2+3)^2}}$, and since $17-(4x^2+3)^2=8(1-3x^2-2x^4)$, this reduces to $\tfrac{x}{\sqrt{1-3x^2-2x^4}}$, matching the integrand; this was also confirmed numerically at $x=0.5$, where both sides equal $\sqrt2\approx1.4142$.