Evaluate the integral: I = integral sqrt(4x²+x) dx

Evaluate the integral: $$ I = \int \sqrt{4x^2+x}\,dx. $$

Step 1 · Complete the square

$$ 4x^2+x = 4\left(x+\frac18\right)^2 - \frac1{16}. $$

So:

$$ \sqrt{4x^2+x} = 2\sqrt{\left(x+\frac18\right)^2-\frac{1}{64}}. $$

Step 2 · Substitute

Let $u=x+\tfrac18$, $du=dx$, $a=\tfrac18$:

$$ I = 2\int \sqrt{u^2-a^2}\,du. $$

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 and simplify

With $u=x+\tfrac18$ and $u^2-a^2=x^2+\tfrac x4=\tfrac14(4x^2+x)$, so $\sqrt{u^2-a^2}=\tfrac12\sqrt{4x^2+x}$:

$$ I = \left(x+\frac18\right)\cdot\frac{\sqrt{4x^2+x}}{2} - \frac1{64}\ln\left|x+\frac18+\frac{\sqrt{4x^2+x}}{2}\right| + C. $$

Multiplying inside the log by $8$ and absorbing the resulting constant into $C$:

$$ I = \frac{8x+1}{16}\sqrt{4x^2+x} - \frac1{64}\ln\left|8x+1+4\sqrt{4x^2+x}\right| + C. $$

Summary Table

Additional Notes

• Correction: the source's stated final answer, $\tfrac{2+x}2\sqrt{4x+x^2}-2\ln|x+2+\sqrt{4x+x^2}|$, is the antiderivative of a different integral, $\int\sqrt{x^2+4x}\,dx$ (note $4x+x^2$, not $4x^2+x$) — the source appears to have solved the wrong problem partway through. The radicand here, $4x^2+x=x(4x+1)$, is a genuinely different quadratic in $x$, requiring its own completing-the-square computation as done above.
• The domain requires $4x^2+x\geq0$, i.e. $x\leq-\tfrac14$ or $x\geq0$.
• The result was verified numerically: differentiating $\tfrac{8x+1}{16}\sqrt{4x^2+x}-\tfrac1{64}\ln|8x+1+4\sqrt{4x^2+x}|$ at $x=1$ gives $\approx2.24$, matching $\sqrt{4(1)^2+1}=\sqrt5\approx2.236$.