Evaluate the integral: I = integral
Evaluate the integral: $$ I = \int \frac{\sqrt{x+1}+2}{(x+1)^2-\sqrt{x+1}}\,dx. $$
Explanation of Method
Substituting $u=\sqrt{x+1}$ turns the denominator into $u(u^3-1)$, and factoring $u^3-1=(u-1)(u^2+u+1)$ reduces the integral to standard partial fractions over a linear and an irreducible quadratic factor.
- Substitute $u=\sqrt{x+1}$.
- Factor the denominator.
- Decompose into partial fractions.
- Integrate and back-substitute.
Step 1 · Substitute
Let $u=\sqrt{x+1}$, so $x=u^2-1$, $dx=2u\,du$, and $(x+1)^2-\sqrt{x+1}=u^4-u=u(u^3-1)$:
$$ I = \int\frac{u+2}{u(u^3-1)}\cdot2u\,du = 2\int\frac{u+2}{u^3-1}\,du. $$
Step 2 · Factor and decompose
$$ u^3-1=(u-1)\left(u^2+u+1\right), \qquad \frac{u+2}{(u-1)(u^2+u+1)} = \frac1{u-1} - \frac{u+1}{u^2+u+1}. $$
(Setting $u=1$ gives the numerator coefficient $A=1$ over $u-1$; matching the remaining coefficients gives $B=-1,\ C=-1$ for the quadratic term.)
Step 3 · Integrate
$$ 2\int\left(\frac1{u-1}-\frac{u+1}{u^2+u+1}\right)du = 2\ln|u-1| - \ln\left(u^2+u+1\right) - \frac2{\sqrt3}\arctan\left(\frac{2u+1}{\sqrt3}\right) + C. $$
Step 4 · Back-substitute
Since $u^2+u+1=(x+1)+\sqrt{x+1}+1=x+\sqrt{x+1}+2$:
$$ I = 2\ln\left|\sqrt{x+1}-1\right| - \ln\left(x+\sqrt{x+1}+2\right) - \frac{2\sqrt3}3\arctan\left(\frac{2\sqrt{x+1}+1}{\sqrt3}\right) + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ (over $u-1$) | $1$ |
| $B$ (coeff. of $u$) | $-1$ |
| $C$ (constant) | $-1$ |
Additional Notes
- The domain requires $x\geq-1,\ x\neq0$.
- The result was verified numerically at $x=3$ (where $u=2$): the chain-rule derivative of the boxed expression evaluates to $2/7$, matching the integrand $(2+2)/[16-2]=4/14=2/7$ there.
Result
The solution to the integral is: