Evaluate the integral: I = integral dx/(sqrt(4x²+4x-3))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{4x^2+4x-3}}. $$
Explanation of Method
Completing the square converts the radical into the standard $\sqrt{u^2-a^2}$ form, resolved by the logarithmic formula $\int du/\sqrt{u^2-a^2}=\ln|u+\sqrt{u^2-a^2}|$.
- Complete the square for $4x^2+4x-3$.
- Substitute $u=x+\tfrac12$.
- Apply the standard logarithmic formula.
- Substitute back, carefully tracking the leading factor of $2$.
Step 1 · Complete the square
$$ 4x^2+4x-3 = 4\left(x+\frac12\right)^2 - 4. $$
So:
$$ \sqrt{4x^2+4x-3} = 2\sqrt{\left(x+\frac12\right)^2-1}. $$
Step 2 · Substitute
Let $u=x+\tfrac12$, $du=dx$, $a=1$:
$$ I = \frac12\int \frac{du}{\sqrt{u^2-a^2}}. $$
Step 3 · Apply the standard logarithmic formula
$$ I = \frac12\ln\left|u+\sqrt{u^2-1}\right| + C. $$
Step 4 · Substitute back, tracking the leading factor
With $u=x+\tfrac12$ and $u^2-1=x^2+x+\tfrac14-1=x^2+x-\tfrac34=\tfrac14(4x^2+4x-3)$, so $\sqrt{u^2-1}=\sqrt{4x^2+4x-3}/2$ — not simply $\sqrt{4x^2+4x-3}$:
$$ I = \frac12\ln\left|x+\frac12+\frac{\sqrt{4x^2+4x-3}}{2}\right| + C. $$
Multiplying inside the log by $2$ and absorbing the resulting constant into $C$:
$$ I = \frac12\ln\left|2x+1+\sqrt{4x^2+4x-3}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Completed square | $4(x+\tfrac12)^2-4$ |
| $u,\,a$ | $x+\tfrac12,\ 1$ |
Additional Notes
- Correction: the source wrote $\sqrt{(x+1/2)^2-1}=\sqrt{4x^2+4x-3}$ directly, but this drops the leading factor of $2$ — the correct relation is $\sqrt{(x+1/2)^2-1}=\sqrt{4x^2+4x-3}/2$. A numerical check at $x=1$ (source's answer would give derivative $\approx0.493$ there, versus the true value $1/\sqrt5\approx0.447$) confirms the error.
- The domain requires $4x^2+4x-3\geq0$, i.e. $x\leq-\tfrac32$ or $x\geq\tfrac12$.
- The result was verified numerically: differentiating $\tfrac12\ln|2x+1+\sqrt{4x^2+4x-3}|$ at $x=1$ gives $\approx0.447$, matching $1/\sqrt5$.
Result
The solution to the integral is: