Evaluate the integral: I = integral dx/(2x² + 1)
Evaluate the integral: $$ I = \int \frac{dx}{2x^2 + 1}. $$
Explanation of Method
Factoring out the leading coefficient of $x^2$ turns the denominator into the standard form $x^2+a^2$, after which the general formula $\int \frac{dx}{x^2+a^2}=\frac1a\arctan\left(\frac{x}{a}\right)$ applies directly, with $a$ possibly irrational.
- Factor the leading coefficient out of the quadratic denominator.
- Identify $a^2$ and hence $a$ from the standard form.
- Apply the standard arctangent integral formula.
- Simplify the resulting expression and add the constant of integration.
Step 1 · Factor the denominator
We can factor the denominator to simplify the expression. The denominator is $2x^2+1$, which can be written as:
$$ 2x^2 + 1 = 2\left(x^2 + \frac{1}{2}\right). $$
Thus, the integral becomes:
$$ I = \int \frac{dx}{2\left(x^2+\frac{1}{2}\right)}. $$
Now, factor out the constant $\dfrac{1}{2}$:
$$ I = \frac{1}{2}\int \frac{dx}{x^2+\frac{1}{2}}. $$
Step 2 · Use a standard integral form
We recognize that the integral $\displaystyle \int \frac{dx}{x^2+a^2}$ has a standard solution, which is:
$$ \int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right). $$
In our case, $a^2 = \dfrac{1}{2}$, so $a = \dfrac{1}{\sqrt{2}}$. Thus, the integral becomes:
$$ I = \frac{1}{2}\cdot \frac{1}{\frac{1}{\sqrt{2}}}\arctan\left(\frac{x}{\frac{1}{\sqrt{2}}}\right) + C. $$
Step 3 · Simplify the expression
Simplifying the expression, we get:
$$ I = \frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}\,x\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Factored form | $2\left(x^2+\tfrac12\right)$ |
| $a$ | $1/\sqrt{2}$ |
| Antiderivative | $\tfrac{\sqrt2}{2}\arctan(\sqrt2\,x)$ |
Additional Notes
- The integrand is defined for all real $x$, since $2x^2+1 \geq 1 > 0$ always.
- The coefficient $\dfrac{\sqrt2}{2}$ simplifies from $\dfrac{1}{2}\cdot\sqrt2 = \dfrac{\sqrt2}{2}$; equivalently, $\dfrac{1}{\sqrt2}$, since $\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$.
- The result can be checked by differentiating $\dfrac{\sqrt2}{2}\arctan(\sqrt2\,x)$ with the chain rule and confirming it returns $\dfrac{1}{2x^2+1}$.
Result
The solution to the integral is: