Evaluate the integral: I = integral
Evaluate the integral: $$ I = \int \frac{3x+1}{\sqrt{2x^2-x+1}}\,dx. $$
Explanation of Method
Writing the numerator as a multiple of the radicand's derivative plus a constant splits the integral into a direct square-root piece and a standard logarithmic piece — since $2x^2-x+1$ is always positive, this calls for the $\int du/\sqrt{u^2+a^2}$ formula, not arcsine.
- Rewrite $3x+1=\tfrac34(4x-1)+\tfrac74$.
- Split into a derivative-matching piece and a constant piece.
- Integrate the first piece directly.
- Complete the square and integrate the second piece using the logarithmic formula.
Step 1 · Rewrite the numerator
Since $\dfrac{d}{dx}(2x^2-x+1)=4x-1$, write:
$$ 3x+1 = \frac34(4x-1) + \frac74. $$
Step 2 · Split the integral
$$ I = \frac34\int \frac{4x-1}{\sqrt{2x^2-x+1}}\,dx + \frac74\int \frac{dx}{\sqrt{2x^2-x+1}}. $$
Step 3 · Integrate the first piece
Let $v=2x^2-x+1$, $dv=(4x-1)\,dx$:
$$ \frac34\int \frac{4x-1}{\sqrt{2x^2-x+1}}\,dx = \frac34\int \frac{dv}{\sqrt v} = \frac32\sqrt v = \frac32\sqrt{2x^2-x+1}. $$
Step 4 · Complete the square for the second piece
$$ 2x^2-x+1 = 2\left(x-\frac14\right)^2 + \frac78, \qquad \sqrt{2x^2-x+1} = \sqrt2\,\sqrt{\left(x-\frac14\right)^2+\frac7{16}}. $$
Let $w=x-\tfrac14$, $a^2=\tfrac7{16}$:
$$ \int \frac{dx}{\sqrt{2x^2-x+1}} = \frac{1}{\sqrt2}\int \frac{dw}{\sqrt{w^2+a^2}} = \frac{1}{\sqrt2}\ln\left|w+\sqrt{w^2+a^2}\right| + C. $$
Since $w^2+a^2=x^2-\tfrac x2+\tfrac12=\tfrac12(2x^2-x+1)$, so $\sqrt{w^2+a^2}=\sqrt{2x^2-x+1}/\sqrt2$, multiplying inside the log by $4\sqrt2$ and absorbing the constant gives:
$$ \int \frac{dx}{\sqrt{2x^2-x+1}} = \frac{1}{\sqrt2}\ln\left|4x-1+2\sqrt2\sqrt{2x^2-x+1}\right| + C. $$
Step 5 · Combine
$$ I = \frac32\sqrt{2x^2-x+1} + \frac{7}{4\sqrt2}\ln\left|4x-1+2\sqrt2\sqrt{2x^2-x+1}\right| + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $(4x-1)/\sqrt{2x^2-x+1}$ | $2\sqrt{2x^2-x+1}$ |
| $1/\sqrt{2x^2-x+1}$ | $\tfrac1{\sqrt2}\ln|4x-1+2\sqrt2\sqrt{2x^2-x+1}|$ |
Additional Notes
- Correction: the source attempted to force an $\arcsin$ formula onto $\int dx/\sqrt{2x^2-x+1}$, but since $2x^2-x+1>0$ for all real $x$ (discriminant $1-8=-7$), the radicand never has the bounded $a^2-u^2$ form that arcsine requires — it is always an unbounded $u^2+a^2$ form, which needs the logarithmic (equivalently, inverse hyperbolic sine) antiderivative. A numerical check confirms the error: at $x=1$, the source's expression $(4x-1)/\sqrt7=3/\sqrt7\approx1.13$ falls outside $\arcsin$'s domain of $[-1,1]$, so the source's stated answer is undefined exactly where the original integrand is perfectly well-behaved.
- The integrand is defined for all real $x$.
- The result was verified directly by differentiation: writing the product-rule and log-rule derivatives of $\tfrac32\sqrt{2x^2-x+1}+\tfrac{7\sqrt2}8\ln|4x-1+2\sqrt2\sqrt{2x^2-x+1}|$ shows the log term's derivative simplifies to $\tfrac{7}{4\sqrt{2x^2-x+1}}$ once the $(4x-1+2\sqrt2\sqrt{2x^2-x+1})$ factor cancels between numerator and denominator, and adding this to the first term's derivative $\tfrac{3(4x-1)}{4\sqrt{2x^2-x+1}}$ gives exactly $\tfrac{3x+1}{\sqrt{2x^2-x+1}}$, matching the integrand.
Result
The solution to the integral is: