Evaluate the integral: I = integral sqrt(4x²-2x+1) dx
Evaluate the integral: $$ I = \int \sqrt{4x^2-2x+1}\,dx. $$
Explanation of Method
Completing the square converts the radicand into the standard $u^2+a^2$ form, resolved by the well-known formula $\int\sqrt{u^2+a^2}\,du=\tfrac u2\sqrt{u^2+a^2}+\tfrac{a^2}2\ln|u+\sqrt{u^2+a^2}|$.
- Complete the square for $4x^2-2x+1$.
- Substitute $u=x-\tfrac14$.
- Apply the standard formula for $\int\sqrt{u^2+a^2}\,du$.
- Substitute back.
Step 1 · Complete the square
$$ 4x^2-2x+1 = 4\left(x-\frac14\right)^2 + \frac34. $$
Dividing by $4$, this radicand equals $4$ times $\left(x-\tfrac14\right)^2+\tfrac3{16}$, so:
$$ \sqrt{4x^2-2x+1} = 2\sqrt{x^2-\frac x2+\frac14}, \qquad \text{where } x^2-\frac x2+\frac14 = \left(x-\frac14\right)^2+\frac3{16}. $$
Step 2 · Substitute
Let $u=x-\tfrac14$, $du=dx$, $a^2=\tfrac3{16}$. Since $\sqrt{4x^2-2x+1}=2\sqrt{u^2+a^2}$:
$$ I = 2\int \sqrt{u^2+a^2}\,du. $$
Step 3 · Apply the standard formula
$$ I = 2\left[\frac u2\sqrt{u^2+a^2} + \frac{a^2}2\ln\left|u+\sqrt{u^2+a^2}\right|\right] + C = u\sqrt{u^2+a^2} + a^2\ln\left|u+\sqrt{u^2+a^2}\right| + C. $$
Step 4 · Substitute back
With $u=x-\tfrac14$ and $\sqrt{u^2+a^2}=\sqrt{x^2-\tfrac x2+\tfrac14}=\tfrac12\sqrt{4x^2-2x+1}$:
$$ I = \frac{4x-1}{4}\sqrt{x^2-\frac x2+\frac14} + \frac3{16}\ln\left|x-\frac14+\sqrt{x^2-\frac x2+\frac14}\right| + C, $$
which is equivalent, using $\sqrt{x^2-\tfrac x2+\tfrac14}=\tfrac12\sqrt{4x^2-2x+1}$, to the cleaner form:
$$ I = \frac{4x-1}{8}\sqrt{4x^2-2x+1} + \frac3{16}\ln\left|4x-1+2\sqrt{4x^2-2x+1}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Completed square | $4(x-\tfrac14)^2+\tfrac34$ |
| $u,\,a^2$ | $x-\tfrac14,\ \tfrac3{16}$ |
Additional Notes
- The integrand is defined for all real $x$, since $4x^2-2x+1>0$ everywhere (discriminant $4-16=-12$).
- Unlike the $\sqrt{\text{quadratic}}$-in-a-denominator problems (197–199), here the leading-coefficient tracking works out cleanly because both sides of the identity $\sqrt{4x^2-2x+1}=2\sqrt{(x-1/4)^2+3/16}$ were applied consistently throughout — this was verified symbolically by direct differentiation rather than left as a numerical check.
- The result was checked by differentiating the $u,a$-form $I=u\sqrt{u^2+a^2}+a^2\ln|u+\sqrt{u^2+a^2}|$ directly: the product rule gives $\sqrt{u^2+a^2}+u^2/\sqrt{u^2+a^2}$, the log rule gives $a^2/\sqrt{u^2+a^2}$ (after the $(u+\sqrt{u^2+a^2})$ factors cancel), and summing gives $[(u^2+a^2)+u^2+a^2]/\sqrt{u^2+a^2}$... more directly, $\sqrt{u^2+a^2}+(u^2+a^2)/\sqrt{u^2+a^2}=2\sqrt{u^2+a^2}$, which equals $\sqrt{4x^2-2x+1}$ exactly, matching the integrand.
Result
The solution to the integral is: