Evaluate the integral: I = integral (x³)/((1+x)
Evaluate the integral: $$ I = \int \frac{x^3}{(1+x)\sqrt{1+2x-x^2}}\,dx. $$
Explanation of Method
Shifting to $u=x-1$ centers the quadratic as $2-u^2$; polynomial division of the resulting numerator by $u+2$ splits the integral into a standard $\int(u^2+u+1)/\sqrt{2-u^2}\,du$ piece and a $\int du/[(u+2)\sqrt{2-u^2}]$ piece, the latter resolved via a Weierstrass ($t=\tan(\theta/2)$) substitution.
- Substitute $u=x-1$ and divide.
- Integrate the polynomial-over-radical piece.
- Integrate the remaining piece via trig + Weierstrass substitution.
- Combine and back-substitute.
Step 1 · Substitute and divide
Let $u=x-1$, so $1+2x-x^2=2-u^2$, $1+x=u+2$, $x^3=(u+1)^3=u^3+3u^2+3u+1$. Polynomial division gives $u^3+3u^2+3u+1=(u+2)(u^2+u+1)-1$:
$$ I = \int\frac{u^2+u+1}{\sqrt{2-u^2}}\,du - \int\frac{du}{(u+2)\sqrt{2-u^2}}. $$
Step 2 · First piece
Using the standard forms for $\int u^2/\sqrt{2-u^2}\,du$, $\int u/\sqrt{2-u^2}\,du$, $\int du/\sqrt{2-u^2}$:
$$ \int\frac{u^2+u+1}{\sqrt{2-u^2}}\,du = 2\arcsin\left(\frac u{\sqrt2}\right) - \frac{u+2}2\sqrt{2-u^2} + C_1. $$
Step 3 · Second piece via $u=\sqrt2\sin\theta$, then Weierstrass
Substituting $u=\sqrt2\sin\theta$ reduces $\int du/[(u+2)\sqrt{2-u^2}]$ to $\int d\theta/(\sqrt2\sin\theta+2)$; the Weierstrass substitution $t=\tan(\theta/2)$ (with $\sin\theta=2t/(1+t^2)$) then gives:
$$ \int\frac{d\theta}{\sqrt2\sin\theta+2} = \int\frac{dt}{t^2+\sqrt2t+1} = \sqrt2\arctan\left(\sqrt2t+1\right) + C_2, $$
where $t=\tan(\theta/2)=\dfrac u{\sqrt2+\sqrt{2-u^2}}$ (from $\sin\theta=u/\sqrt2$).
Step 4 · Combine and back-substitute
$$ I = 2\arcsin\left(\frac u{\sqrt2}\right) - \frac{u+2}2\sqrt{2-u^2} - \sqrt2\arctan\left(\sqrt2\cdot\frac u{\sqrt2+\sqrt{2-u^2}}+1\right) + C, $$
and with $u=x-1$, $2-u^2=1+2x-x^2$:
Summary Table
| Piece | Antiderivative |
|---|---|
| $(u^2+u+1)/\sqrt{2-u^2}$ | $2\arcsin(u/\sqrt2)-\tfrac{u+2}2\sqrt{2-u^2}$ |
| $-1/[(u+2)\sqrt{2-u^2}]$ | $-\sqrt2\arctan(\sqrt2t+1)$, $t=u/(\sqrt2+\sqrt{2-u^2})$ |
Additional Notes
- Major rewrite: the source's Step 6 asserted the closed form for $I_2=\int du/[(u+2)\sqrt{2-u^2}]$ was "looked up in integral tables" without stating or verifying the actual formula, then combined it inconsistently in Step 7. The full derivation above (trig substitution followed by a Weierstrass substitution) was carried out explicitly and verified independently, term by term.
- The domain is $1-\sqrt2\leq x\leq1+\sqrt2$.
- The result was verified by differentiation at $x=0,\,0.5,\,-0.3,\,1.5$: the derivative matches the integrand $x^3/[(1+x)\sqrt{1+2x-x^2}]$ exactly at all four points.
Result
The solution to the integral is: