Evaluate the integral: I = integral (1- sqrt(x+1))/(1+
Evaluate the integral: $$ I = \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}\,dx. $$
Explanation of Method
Substituting $u=(x+1)^{1/6}$ rationalizes both radicals; long-dividing the resulting $u^5,u^8$ terms by $1+u^2$ separates a polynomial piece from a derivative-matching log/arctangent piece.
- Substitute $u=(x+1)^{1/6}$.
- Divide $u^5$ and $u^8$ each by $(1+u^2)$.
- Integrate the combined expression.
- Back-substitute.
Step 1 · Substitute
Let $u=(x+1)^{1/6}$, so $x=u^6-1$, $dx=6u^5\,du$, $\sqrt{x+1}=u^3$, $\sqrt[3]{x+1}=u^2$:
$$ I = \int\frac{1-u^3}{1+u^2}\cdot6u^5\,du = 6\int\frac{u^5-u^8}{1+u^2}\,du. $$
Step 2 · Divide each power
$$ \frac{u^5}{1+u^2}=u^3-u+\frac u{1+u^2}, \qquad \frac{u^8}{1+u^2}=u^6-u^4+u^2-1+\frac1{1+u^2}. $$
Subtracting:
$$ \frac{u^5-u^8}{1+u^2} = -u^6+u^4+u^3-u^2-u+1+\frac{u-1}{1+u^2}. $$
Step 3 · Integrate
$$ 6\int\left(-u^6+u^4+u^3-u^2-u+1+\frac{u-1}{1+u^2}\right)du $$
$$ = -\frac{6u^7}7+\frac{6u^5}5+\frac{3u^4}2-2u^3-3u^2+6u+3\ln\left(1+u^2\right)-6\arctan u + C. $$
Step 4 · Back-substitute
$$ I = -\frac{6(x+1)^{7/6}}7 + \frac{6(x+1)^{5/6}}5 + \frac{3(x+1)^{2/3}}2 - 2\sqrt{x+1} - 3\sqrt[3]{x+1} + 6\sqrt[6]{x+1} + 3\ln\left(1+\sqrt[3]{x+1}\right) - 6\arctan\left(\sqrt[6]{x+1}\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $-u^6+u^4+u^3-u^2-u+1$ | degree-7 polynomial (shown above) |
| $(u-1)/(1+u^2)$ | $\tfrac12\ln(1+u^2)-\arctan u$ |
Additional Notes
- The domain requires $x\geq-1$.
- The result was verified numerically at $x=63$ (where $u=2$): the chain-rule derivative evaluates to $-1.4$, matching the integrand $(1-8)/(1+4)=-7/5=-1.4$ there.
Result
The solution to the integral is: