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 $u+1=\sqrt2\sec\theta$.
- 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
$$ u^2+2u-1 = (u+1)^2-2, \qquad I = \frac12\int \sqrt{(u+1)^2-2}\,du. $$
Step 3 · Secant substitution
Let $u+1=\sqrt2\sec\theta$, $du=\sqrt2\sec\theta\tan\theta\,d\theta$, so $\sqrt{(u+1)^2-2}=\sqrt2\tan\theta$:
$$ I = \frac12\int \sqrt2\tan\theta\cdot\sqrt2\sec\theta\tan\theta\,d\theta = \int \tan^2\theta\sec\theta\,d\theta. $$
Step 4 · Reduce using $\tan^2\theta=\sec^2\theta-1$
$$ \int \tan^2\theta\sec\theta\,d\theta = \int \left(\sec^3\theta-\sec\theta\right) d\theta = \frac12\sec\theta\tan\theta - \frac12\ln\left|\sec\theta+\tan\theta\right| + C, $$
using the standard results $\int\sec^3\theta\,d\theta=\tfrac12\sec\theta\tan\theta+\tfrac12\ln|\sec\theta+\tan\theta|$ and $\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|$.
Step 5 · Substitute back
With $\sec\theta=\dfrac{u+1}{\sqrt2}=\dfrac{x^2+1}{\sqrt2}$ and $\tan\theta=\dfrac{\sqrt{(u+1)^2-2}}{\sqrt2}=\dfrac{\sqrt{x^4+2x^2-1}}{\sqrt2}$:
$$ \sec\theta\tan\theta = \frac{(x^2+1)\sqrt{x^4+2x^2-1}}{2}, \qquad \sec\theta+\tan\theta = \frac{x^2+1+\sqrt{x^4+2x^2-1}}{\sqrt2}. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u=x^2$, then $u+1=\sqrt2\sec\theta$ |
| 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 is a two-layer substitution: first $u=x^2$ reduces the quartic-under-radical problem to a quadratic-under-radical one, then the secant substitution handles that quadratic exactly as in the $\sqrt{u^2-a^2}$ pattern from Problem 182.
- The result was verified directly by differentiation: writing $F(x)=\tfrac{(x^2+1)\sqrt s}{4}-\tfrac12\ln\left|\tfrac{x^2+1+\sqrt s}{\sqrt2}\right|$ with $s=x^4+2x^2-1$, the product-rule and log-rule derivatives combine (using $s+(x^2+1)^2=2x^4+4x^2=2(x^4+2x^2)$) to give exactly $x\sqrt s$, matching the integrand.
Result
The solution to the integral is: