Evaluate the integral: I = integral dx/(sqrt(2x²+3x))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{2x^2+3x}}. $$
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 $2x^2+3x$.
- Substitute $u=x+\tfrac34$.
- Apply the standard logarithmic formula.
- Substitute back and simplify.
Step 1 · Complete the square
$$ 2x^2+3x = 2\left(x^2+\frac32x\right) = 2\left[\left(x+\frac34\right)^2-\frac9{16}\right] = 2\left(x+\frac34\right)^2 - \frac98. $$
So:
$$ \sqrt{2x^2+3x} = \sqrt2\,\sqrt{\left(x+\frac34\right)^2-\frac9{16}}. $$
Step 2 · Substitute
Let $w=x+\tfrac34$, $dw=dx$, $a=\tfrac34$:
$$ I = \frac{1}{\sqrt2}\int \frac{dw}{\sqrt{w^2-a^2}}. $$
Step 3 · Apply the standard logarithmic formula
$$ I = \frac{1}{\sqrt2}\ln\left|w+\sqrt{w^2-a^2}\right| + C. $$
Step 4 · Substitute back and simplify
With $w=x+\tfrac34$ and $w^2-a^2=x^2+\tfrac32x=\tfrac12(2x^2+3x)$, so $\sqrt{w^2-a^2}=\sqrt{2x^2+3x}/\sqrt2$:
$$ I = \frac{1}{\sqrt2}\ln\left|x+\frac34+\frac{\sqrt{2x^2+3x}}{\sqrt2}\right| + C. $$
Multiplying inside the log by $4$ and absorbing the resulting constant into $C$ gives the cleaner equivalent form:
$$ I = \frac{1}{\sqrt2}\ln\left|4x+3+2\sqrt2\sqrt{2x^2+3x}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Completed square | $2(x+\tfrac34)^2-\tfrac98$ |
| $w,\,a$ | $x+\tfrac34,\ \tfrac34$ |
| Domain | $x\leq-\tfrac32$ or $x\geq0$ |
Additional Notes
- Correction: the source's substitution $x=-\tfrac32u$ eventually required $\int du/\sqrt{u(u-1)}$, but the source evaluated this as $\ln|\sqrt u+\sqrt{u-1}|$ — missing a factor of $2$ (the correct value, following the same derivation used in Problem 193, is $2\ln|\sqrt u+\sqrt{u-1}|$). This dropped factor propagated into the final coefficient, and the resulting answer also only covered the $x\leq-3/2$ branch of the domain, not $x\geq0$. Completing the square directly, as done here, avoids both issues and covers the full domain.
- The domain requires $2x^2+3x\geq0$, i.e. $x\leq-\tfrac32$ or $x\geq0$.
- The result can be checked by differentiating $\tfrac{1}{\sqrt2}\ln|4x+3+2\sqrt2\sqrt{2x^2+3x}|$: the derivative of the log's argument is $4+2\sqrt2\cdot\tfrac{4x+3}{2\sqrt{2x^2+3x}}=4+\sqrt2\cdot\tfrac{4x+3}{\sqrt{2x^2+3x}}$; combining over $\sqrt{2x^2+3x}$ and simplifying the resulting expression (multiplying numerator and denominator appropriately) reduces to $\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt{2x^2+3x}}\cdot\sqrt2=\tfrac{1}{\sqrt{2x^2+3x}}$, matching the integrand; this was confirmed numerically at $x=1$.
Result
The solution to the integral is: