Evaluate the integral (for x>0): I = integral dx/([1+
Evaluate the integral (for $x>0$): $$ I = \int \frac{dx}{\left[1+\sqrt{x(1+x)}\right]^2}. $$
Explanation of Method
Setting $w=\sqrt{x(1+x)}$ and solving for $x$ in terms of $w$ (using $x^2+x-w^2=0$) turns the integral into a rational-times-radical form in $w$; a further reciprocal substitution $t=1/(1+w)$ rationalizes the radical completely, leaving a standard $\int dt/\sqrt{\text{quadratic}}$-type integral.
- Substitute $w=\sqrt{x(1+x)}$.
- Substitute $t=1/(1+w)$.
- Split and integrate the resulting rational-radical pieces.
- Back-substitute twice.
Step 1 · Substitute $w=\sqrt{x(1+x)}$
From $w^2=x^2+x$, solving the quadratic in $x$ (positive root, $x>0$) gives $x=\tfrac{-1+\sqrt{1+4w^2}}2$, so $1+2x=\sqrt{1+4w^2}$ and:
$$ dx = \frac{2w\,dw}{\sqrt{1+4w^2}}, \qquad I = \int\frac{2w\,dw}{(1+w)^2\sqrt{1+4w^2}}. $$
Step 2 · Split via $w=(1+w)-1$
$$ \frac{2w}{(1+w)^2} = \frac{2}{1+w} - \frac{2}{(1+w)^2}. $$
$$ I = 2\int\frac{dw}{(1+w)\sqrt{1+4w^2}} - 2\int\frac{dw}{(1+w)^2\sqrt{1+4w^2}}. $$
Step 3 · Reciprocal substitution $t=1/(1+w)$
With $t=1/(1+w)$, $w=(1-t)/t$, $dw=-dt/t^2$, one finds $1+4w^2=(5t^2-8t+4)/t^2$, and both integrals reduce to standard forms in $t$:
$$ \int\frac{dw}{(1+w)\sqrt{1+4w^2}} = -\int\frac{dt}{\sqrt{5t^2-8t+4}} = -\frac{\sqrt5}5\,\text{arcsinh}\left(\frac{5t}2-2\right), $$
$$ \int\frac{dw}{(1+w)^2\sqrt{1+4w^2}} = -\int\frac{t\,dt}{\sqrt{5t^2-8t+4}} = -\frac{\sqrt{5t^2-8t+4}}5 - \frac{4\sqrt5}{25}\,\text{arcsinh}\left(\frac{5t}2-2\right). $$
Step 4 · Combine and back-substitute
$$ I = \frac25\sqrt{5t^2-8t+4} - \frac{2\sqrt5}{25}\,\text{arcsinh}\left(\frac{5t}2-2\right) + C. $$
With $t=1/(1+w)$: $\sqrt{5t^2-8t+4}=\sqrt{1+4w^2}/(1+w)=(2x+1)/(1+w)$, and $\tfrac{5t}2-2=\tfrac{1-4w}{2(1+w)}$:
Summary Table
| Substitution | Effect |
|---|---|
| $w=\sqrt{x(1+x)}$ | reduces to $\int2w\,dw/[(1+w)^2\sqrt{1+4w^2}]$ |
| $t=1/(1+w)$ | reduces both pieces to standard $\int dt/\sqrt{5t^2-8t+4}$-type forms |
Additional Notes
- Major rewrite: the source's Steps 3–4 abandoned a valid substitution chain midway ("this becomes complex... let's try a different approach") and its final Step 6 asserted a boxed answer, $\sqrt{x(1+x)}/[1+x+\sqrt{x(1+x)}]$, by pattern-guessing rather than deriving it. That answer fails numerically (derivative $\approx0.0607$ vs. true integrand $\approx0.1716$ at $x=1$). The boxed result below completes the substitution chain the source itself started in Step 5 and was verified independently.
- The domain is $x>0$.
- The result was verified by differentiation at $x=1,2,3$: the derivative matches the integrand $1/[1+\sqrt{x(1+x)}]^2$ exactly at all three points.
Result
The solution to the integral is, with $w=\sqrt{x(1+x)}$: