AP Calculus AB and BC

Integral of 1/(x^2+4): Answer, Proof, and Steps

Q: Why does a 1/2 appear out front?

Differentiating \arctan\frac{x}{2} gives \frac{1}{2}\cdot\frac{1}{1+\frac{x^2}{4}} = \frac{2}{x^2+4}. The inner derivative contributes \frac{1}{a}, and clearing the compound fraction contributes a^2, so the derivative overshoots the integrand by a factor of a. Dividing by a = 2 is where the \frac{1}{2} comes from, and in general the coefficient is \frac{1}{a}.

The integral of 1/(x^2+4) is (1/2)arctan(x/2) + C. The integrand matches the standard form 1/(x^2 + a^2) with a = 2, whose antiderivative is (1/a)arctan(x/a). Differentiating (1/2)arctan(x/2) returns 1/(x^2+4), confirming the result.

\int \frac{1}{x^2+4}\,dx = \frac{1}{2}\arctan\frac{x}{2} + C

Matching the arctangent form

A constant over a sum of squares is the arctangent pattern. The constant term names $a^2$ , so here $a^2 = 4$ and $a = 2$ .

\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C

\int \frac{1}{x^2+4}\,dx = \frac{1}{2}\arctan\frac{x}{2} + C

The formula itself comes from substitution. Put $x = 2u$ , so $dx = 2\,du$ and $x^2 + 4 = 4(u^2+1)$ .

\int \frac{2\,du}{4(u^2+1)} = \frac{1}{2}\int \frac{du}{u^2+1} = \frac{1}{2}\arctan u + C = \frac{1}{2}\arctan\frac{x}{2} + C

Checking by differentiating

Differentiate with $\frac{d}{dx}\arctan u = \frac{u'}{1+u^2}$ , where $u = \frac{x}{2}$ and $u' = \frac{1}{2}$ .

\frac{d}{dx}\left(\frac{1}{2}\arctan\frac{x}{2}\right) = \frac{1}{2}\cdot\frac{\frac{1}{2}}{1+\frac{x^2}{4}} = \frac{1}{4}\cdot\frac{4}{4+x^2} = \frac{1}{x^2+4}

Multiplying the inner fraction top and bottom by $4$ clears the compound fraction and returns the integrand.

The mistakes students make

Answering $\arctan\frac{x}{2}$ with no $\frac{1}{2}$ in front. Differentiating that gives $\frac{2}{x^2+4}$ , twice the integrand.
Using $a = 4$ instead of $a = 2$ and writing $\frac{1}{4}\arctan\frac{x}{4}$ . The constant in the denominator is $a^2$ , not $a$ .
Answering $\ln(x^2+4)$ . A logarithm appears only when the numerator matches the derivative of the denominator, which would need an $x$ on top.
Confusing this with $\int\frac{1}{\sqrt{4-x^2}}\,dx$ , which is an arcsine problem. A square root and a subtraction signal arcsine, not arctangent.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

Why does a 1/2 appear out front?

Differentiating $\arctan\frac{x}{2}$ gives $\frac{1}{2}\cdot\frac{1}{1+\frac{x^2}{4}} = \frac{2}{x^2+4}$ . The inner derivative contributes $\frac{1}{a}$ , and clearing the compound fraction contributes $a^2$ , so the derivative overshoots the integrand by a factor of $a$ . Dividing by $a = 2$ is where the $\frac{1}{2}$ comes from, and in general the coefficient is $\frac{1}{a}$ .

What is the integral of 1/(x^2+4) from 0 to 2?

$\int_0^2 \frac{1}{x^2+4}\,dx = \left[\frac{1}{2}\arctan\frac{x}{2}\right]_0^2 = \frac{1}{2}\arctan 1 = \frac{\pi}{8}$ .

What changes if the numerator is x instead of 1?

Then the numerator is a constant multiple of the derivative of $x^2+4$ , so substitution with $u = x^2+4$ gives $\int\frac{x}{x^2+4}\,dx = \frac{1}{2}\ln(x^2+4) + C$ . Check the numerator before choosing a form.