Deriving the Inverse Tangent Formula for Quadratics

Why an Inverse Tangent Appears

A quadratic denominator that has no real roots can often be written as a positive or negative constant times a sum of a square and a positive number. The basic model is \(u^2+a^2\), where \(a>0\). Its reciprocal has an antiderivative involving the inverse tangent because the derivative of \(\arctan z\) is \(1/(1+z^2)\).

The important detail is the scale. Replacing \(z\) by \(u/a\) changes both the argument of the inverse tangent and the coefficient in front of it. Keeping track of both factors gives a formula that can then be translated to any irreducible quadratic.

The Standard Inverse-Tangent Formula

We use \(\arctan z\) for the principal inverse tangent, whose values lie in \((-\pi/2,\pi/2)\). Its derivative is \(1/(1+z^2)\). This derivative follows by setting \(y=\arctan z\), so that \(\tan y=z\), and differentiating: \(\sec^2(y)\,dy/dz=1\). Since \(\sec^2(y)=1+\tan^2(y)=1+z^2\), it follows that \(dy/dz=1/(1+z^2)\).

Let \(a>0\). To find an antiderivative with respect to \(u\), scale the input to the inverse tangent by \(1/a\). By the chain rule,

The two factors of \(1/a\) have different sources: one is the coefficient outside \(\arctan\), and the other comes from differentiating \(u/a\). Combining them with the denominator \(1+(u/a)^2\) produces exactly \(u^2+a^2\).

Derivation by Scaling the Variable

The same result can be understood by a change of scale. Set \(t=u/a\), so \(u=at\) and \(du=a\,dt\). Then \(u^2+a^2=a^2(t^2+1)\), and the integral transforms as follows:

Here we used the standard antiderivative \(\int 1/(1+t^2)\,dt=\arctan t+C\), which follows from the inverse-tangent derivative above. This derivation explains why the coefficient is \(1/a\), not \(1/a^2\): the substitution contributes \(du=a\,dt\), leaving one factor \(1/a\) outside the standard integral.

Translating the Formula to a General Quadratic

Consider \(Ax^2+Bx+C\), where \(A\ne 0\). Its discriminant is \(\Delta=B^2-4AC\). The quadratic is irreducible over the real numbers precisely when \(\Delta<0\). Completing the square gives

When \(\Delta<0\), the second term inside the brackets is positive. Define the positive number \(a=\sqrt{4AC-B^2}/(2|A|)\). Then the completed-square form is \(A[(x+B/(2A))^2+a^2]\). The absolute value in the definition of \(a\) ensures \(a>0\), whether \(A\) is positive or negative.

Let \(u=x+B/(2A)\). Since \(du=dx\), the standard formula gives an antiderivative of the reciprocal quadratic:

Here \(C_0\) denotes the arbitrary constant of integration. Substituting the expressions for \(u\) and \(a\) yields a useful formula expressed directly in the original coefficients:

Why the Coefficient Formula Works

To check the compact formula, write \(D=\sqrt{-\Delta}>0\) and \(z=(2Ax+B)/D\). Then \(z'=2A/D\). Differentiating the proposed antiderivative gives

Since \(D^2=-\Delta=4AC-B^2\), the denominator in this last expression simplifies:

Therefore the derivative is \(4A/[4A(Ax^2+Bx+C)]=1/(Ax^2+Bx+C)\). This verifies the formula directly, including when \(A<0\). The signs in the argument and the coefficient work together; the formula does not require a positive leading coefficient.

Worked Examples with General Quadratics

A Common Scaling Mistake

A frequent error is to scale the argument of \(\arctan\) but omit the corresponding coefficient. For example, an incorrect proposed antiderivative for \(1/(u^2+9)\) is \(\arctan(u/3)\). Differentiating exposes the error:

which is three times the required integrand. The correction is to include the outside factor \(1/3\):

When applying this formula to a quadratic, there may be another scale to track: the leading coefficient \(A\), as well as any factor introduced by the substitution for \(u\). Differentiating the final expression is a reliable way to check that all factors have been included.

When the Formula Applies

The inverse-tangent formula in this tutorial requires a positive sum-of-squares parameter \(a\), or, for the general quadratic, a negative discriminant. A quadratic with \(\Delta<0\) has no real roots, and its completed-square constant is positive after the leading coefficient is factored out. If \(\Delta\ge0\), this derivation does not apply: the quadratic cannot be put into the required irreducible form with a positive \(a^2\).

The constant \(C_0\) is arbitrary on any interval where the antiderivative is being considered. In the irreducible case, the denominator is nonzero on all of \(\mathbb R\), so there is no need to split the real line into separate intervals. The formula is useful because it converts a broad class of quadratic denominators into one standard pattern, while making the required condition and all scale factors explicit.