Evaluate the integral: I = integral dx/(x²sqrt(x²+x-1))
Evaluate the integral: $$ I = \int \frac{dx}{x^2\sqrt{x^2+x-1}}. $$
Explanation of Method
The reciprocal substitution $x=1/t$ turns the $x^2$-multiplied radical into $\int t/\sqrt{1+t-t^2}\,dt$, which splits into a derivative-matching piece and a standard arcsine piece.
- Substitute $x=1/t$.
- Simplify the transformed integral.
- Split the numerator $t$ into a derivative-matching piece and a constant piece.
- Integrate each piece and substitute back.
Step 1 · Reciprocal substitution
Let $x=1/t$, $dx=-dt/t^2$ (taking $t>0$, i.e. $x>0$). Then $x^2=1/t^2$ and:
$$ x^2+x-1 = \frac{1+t-t^2}{t^2}, \qquad \sqrt{x^2+x-1} = \frac{\sqrt{1+t-t^2}}{t}. $$
Step 2 · Simplify
$$ I = \int \frac{-dt/t^2}{\dfrac1{t^2}\cdot\dfrac{\sqrt{1+t-t^2}}{t}} = -\int \frac{t\,dt}{\sqrt{1+t-t^2}}. $$
Step 3 · Split the numerator
Since $\dfrac{d}{dt}(1+t-t^2)=1-2t$, write $t=-\tfrac12(1-2t)+\tfrac12$:
$$ -\int \frac{t\,dt}{\sqrt{1+t-t^2}} = \frac12\int \frac{1-2t}{\sqrt{1+t-t^2}}\,dt - \frac12\int \frac{dt}{\sqrt{1+t-t^2}}. $$
The first piece is a direct square-root match: $\tfrac12\int(1-2t)/\sqrt{1+t-t^2}\,dt=\sqrt{1+t-t^2}$. For the second, complete the square:
$$ 1+t-t^2 = \frac54-\left(t-\frac12\right)^2, \qquad \int \frac{dt}{\sqrt{1+t-t^2}} = \arcsin\left(\frac{2t-1}{\sqrt5}\right). $$
So:
$$ I = \sqrt{1+t-t^2} - \frac12\arcsin\left(\frac{2t-1}{\sqrt5}\right) + C. $$
Step 4 · Substitute back
With $t=1/x$ ($x>0$), $\sqrt{1+t-t^2}=\sqrt{x^2+x-1}/x$ and $2t-1=(2-x)/x$:
$$ I = \frac{\sqrt{x^2+x-1}}{x} - \frac12\arcsin\left(\frac{2-x}{x\sqrt5}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $x=1/t$ |
| Reduced integral | $-\int t\,dt/\sqrt{1+t-t^2}$ |
| Domain | $x\geq\tfrac{-1+\sqrt5}{2}$ (positive branch) |
Additional Notes
- Correction: the source tried several substitutions, got bogged down in algebra each time, and ultimately concluded (incorrectly) that this integral "does not simplify to a basic elementary function." That conclusion is wrong — every $\int dx/[x^n\sqrt{ax^2+bx+c}]$-type integral has an elementary closed form, reachable via the reciprocal substitution $x=1/t$ (as demonstrated above), which the source itself set up correctly but then abandoned right before it would have simplified nicely.
- The domain requires $x^2+x-1\geq0$ and $x\neq0$; for the $x>0$ branch this means $x\geq\tfrac{-1+\sqrt5}{2}\approx0.618$.
- The result was verified directly by differentiation (independent of the substitution's sign assumptions): writing $g(x)=\sqrt{x^2+x-1}/x-\tfrac12\arcsin\!\left(\tfrac{2-x}{x\sqrt5}\right)$, the quotient-rule derivative of the first term gives $(2-x)/(2x^2\sqrt{x^2+x-1})$, and the arcsine term's derivative (using $4(x^2+x-1)=5x^2-(2-x)^2$) gives $1/(2x\sqrt{x^2+x-1})$; summing these over the common denominator $2x^2\sqrt{x^2+x-1}$ gives exactly $1/(x^2\sqrt{x^2+x-1})$, matching the integrand.
Result
The solution to the integral is: