Deriving the Logarithmic Form for a Positive-Sign Radicand

Introduction

A reciprocal square root such as \(1/\sqrt{u^2+a^2}\) has no elementary antiderivative that is a power of the radicand. Its antiderivative can instead be written as a logarithm. This form becomes useful when a quadratic under a square root has been completed to a positive multiple of a square plus a positive constant.

The coefficient matters: after completing the square, a change of scale contributes a factor to the antiderivative. We will derive the basic logarithmic formula, extend it to a completed-square radicand, and check the result by differentiation.

The Basic Logarithmic Antiderivative

Here \(C\) denotes an arbitrary constant. The condition \(a>0\) ensures that the radicand is strictly positive and that the expression inside the logarithm never vanishes. In fact, \(\sqrt{u^2+a^2}>|u|\), so \(u+\sqrt{u^2+a^2}>0\) whether \(u\) is positive, negative, or zero.

To see why the formula works, write \(R=\sqrt{u^2+a^2}\). Differentiating \(R\) gives \(R'=u/R\). Therefore, the derivative of the proposed logarithm is

The cancellation is valid because \(u+R>0\); in particular, the denominator is not zero. This verifies the antiderivative directly. The argument also explains the logarithmic form: differentiating the square root produces a numerator involving \(u\), and the logarithm combines that term with the derivative of \(u\) to give the required reciprocal square root.

Adapting the Formula to a Completed Square

Suppose the radicand has the form \(A(x-h)^2+c\), where \(A>0\) and \(c>0\). The positive coefficient \(A\) changes the scale of the squared term, so the basic formula cannot be used by simply replacing \(u\) with \(x-h\). A substitution accounts for that scale.

To obtain the formula, set \(u=\sqrt{A/c}(x-h)\). Then \(dx=\sqrt{c/A}\,du\), and

Consequently, the integral becomes

The basic formula with \(a=1\) gives \(\ln(u+\sqrt{u^2+1})\). Substituting the definition of \(u\) back in, the logarithm’s argument is

Taking the logarithm of this expression introduces the constant \(-\frac12\ln c\), which is absorbed into the arbitrary integration constant. This yields the stated formula.

For the positivity condition, observe that

Thus \(\sqrt A(x-h)+\sqrt{A(x-h)^2+c}>0\) for all real \(x\), including when \(x-h<0\). The original radicand is also positive for every real \(x\), because \(A(x-h)^2\geq 0\) and \(c>0\).

Using the Formula After Completing the Square

A quadratic with a positive leading coefficient can sometimes be rewritten as a positive multiple of a square plus a positive constant. Once it is in that form, identify \(A\), \(h\), and \(c\) before applying the formula. The coefficient \(1/\sqrt A\) is essential.

A Reliable Procedure and a Common Pitfall

For an integral with a reciprocal square root, first rewrite the radicand as \(A(x-h)^2+c\). If both \(A\) and \(c\) are positive, the formula applies on the whole real line. Then use the coefficient \(1/\sqrt A\) and the same \(A\) inside the logarithmic expression. Finally, differentiate the result to check it.

A frequent error is to omit the factor \(1/\sqrt A\). In the first completed-square example, that mistake would produce \(\ln(2x+1+\sqrt q)\) instead of \(\frac12\ln(2x+1+\sqrt q)\). Differentiating the incorrect expression gives

which is twice the required integrand. The correction is to restore the missing factor \(\frac12=1/\sqrt4\). Checking by differentiation is a direct way to catch this scaling error.

Another possible confusion is whether to place absolute-value bars around the logarithm’s argument. In this setting they are unnecessary: when \(c>0\), the square root is strictly larger than the absolute value of the linear term \(\sqrt A(x-h)\). The argument is therefore already positive everywhere. The conditions matter; this formula is specifically for a positive square term and a positive constant term, not for every possible quadratic radicand.

Why the Logarithmic Form Is Useful

The logarithmic expression gives an explicit antiderivative that is easy to verify and preserves the completed-square structure. In practical integration, completing the square identifies the shift \(h\), while the coefficient \(A\) determines the scale factor in front of the logarithm. Keeping those roles separate helps prevent sign and coefficient errors.

The form is also useful for recognizing when this particular formula applies: the radicand must be expressible as \(A(x-h)^2+c\) with \(A,c>0\). The positivity of \(c\) ensures that the square root never vanishes and that the logarithm is defined for every real input. If either condition is not met, this derivation does not establish an antiderivative for that case.