Evaluate the integral: I = integral x sqrt(x^4+2x²-1) dx
Evaluate the integral: $$ I = \int x\sqrt{x^4+2x^2-1}\,dx. $$
Explanation of Method
Substituting $u=x^2$ turns the radicand into a quadratic in $u$; completing the square and applying a secant substitution reduces the integral to the standard $\int\tan^2\theta\sec\theta\,d\theta$ pattern.
- Substitute $u=x^2$.
- Complete the square in $u^2+2u-1$.
- Apply the secant substitution $v=u+1=\sqrt2\sec\theta$.
- Reduce to $\int(\sec^3\theta-\sec\theta)\,d\theta$ and substitute back.
Step 1 · Substitute $u=x^2$
Let $u=x^2$, $du=2x\,dx$:
$$ I = \frac12\int \sqrt{u^2+2u-1}\,du. $$
Step 2 · Complete the square and shift
$$ u^2+2u-1 = (u+1)^2-2. $$
Let $v=u+1$, $dv=du$:
$$ I = \frac12\int \sqrt{v^2-2}\,dv. $$
Step 3 · Secant substitution
Let $v=\sqrt2\sec\theta$, $dv=\sqrt2\sec\theta\tan\theta\,d\theta$, so $\sqrt{v^2-2}=\sqrt2\tan\theta$:
$$ I = \frac12\int \sqrt2\tan\theta\cdot\sqrt2\sec\theta\tan\theta\,d\theta = \int \tan^2\theta\sec\theta\,d\theta. $$
Step 4 · Reduce using $\tan^2\theta=\sec^2\theta-1$
$$ \int \tan^2\theta\sec\theta\,d\theta = \int \left(\sec^3\theta-\sec\theta\right) d\theta = \frac12\sec\theta\tan\theta - \frac12\ln\left|\sec\theta+\tan\theta\right| + C. $$
Step 5 · Substitute back
With $\sec\theta=\dfrac{v}{\sqrt2}$ and $\tan\theta=\dfrac{\sqrt{v^2-2}}{\sqrt2}$:
$$ \sec\theta\tan\theta = \frac{v\sqrt{v^2-2}}{2}, \qquad \sec\theta+\tan\theta = \frac{v+\sqrt{v^2-2}}{\sqrt2}. $$
With $v=u+1=x^2+1$ and $v^2-2=(x^2+1)^2-2=x^4+2x^2-1$:
$$ I = \frac14(x^2+1)\sqrt{x^4+2x^2-1} - \frac12\ln\left|\frac{x^2+1+\sqrt{x^4+2x^2-1}}{\sqrt2}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u=x^2$, then $v=u+1=\sqrt2\sec\theta$ |
| Reduced integral | $\int(\sec^3\theta-\sec\theta)\,d\theta$ |
Additional Notes
- The domain requires $x^4+2x^2-1\geq0$, i.e. $x^2\geq\sqrt2-1$.
- This problem is identical to Problem 201 elsewhere in this collection (the radicand $(x^2+1)^2-2$ is just the expanded $x^4+2x^2-1$ written in completed-square form); both were solved with the same substitution chain and give the same verified answer.
- The result can be checked by differentiation exactly as in Problem 201: writing $s=x^4+2x^2-1$, the derivative of $\tfrac{(x^2+1)\sqrt s}4-\tfrac12\ln\left|\tfrac{x^2+1+\sqrt s}{\sqrt2}\right|$ reduces, using $s+(x^2+1)^2=2x^4+4x^2$, to exactly $x\sqrt s$, matching the integrand.
Result
The solution to the integral is: