Logarithmic Forms from Substitution

When a Denominator Reappears in Its Derivative

Some integrals have a recognizable structure: the numerator is the derivative of the denominator, or a constant multiple of it. For instance, the derivative of \(x^2+1\) is \(2x\), so an integrand such as \(\frac{2x}{x^2+1}\) suggests a logarithm.

The key is to recognize the relationship between the numerator and denominator, not merely that the integrand is a rational expression. When the relationship matches, substitution reduces the integral to the basic form \(\int \frac{1}{u}\,du\). The resulting antiderivative uses the natural logarithm of an absolute value.

Why the Absolute Value Is Necessary

The real natural logarithm \(\ln y\) is defined only when \(y>0\). But a differentiable function \(g(x)\) need not be positive. It may be negative throughout an interval, and the quotient \(\frac{g'(x)}{g(x)}\) is still defined there as long as \(g(x)\ne0\).

The absolute value makes the logarithm real on either kind of interval: if \(g(x)\ne0\), then \(|g(x)|>0\), so \(\ln|g(x)|\) is defined. The function \(g\) cannot change sign within an interval without passing through zero. Thus, the formula applies on any interval on which \(g\) is nonzero, including intervals where \(g\) is negative.

Deriving the Logarithmic Rule

The rule follows from substitution. Let \(u=g(x)\), so that \(du=g'(x)\,dx\). Wherever \(g(x)\ne0\), the integral becomes the integral of \(1/u\). The absolute value in the antiderivative is required because \(u\) can be positive or negative.

The underlying derivative identity can also be checked directly. For \(g(x)>0\), differentiating \(\ln|g(x)|=\ln(g(x))\) gives \(g'(x)/g(x)\). For \(g(x)<0\), \(|g(x)|=-g(x)\), so differentiating \(\ln(-g(x))\) gives \(-g'(x)/(-g(x))=g'(x)/g(x)\). Both cases give the same result.

The condition \(g(x)\ne0\) matters: at a zero of \(g\), the quotient is undefined, and so is \(\ln|g(x)|\). If \(g\) has zeros, apply the rule on intervals that do not contain those zeros. The constant of integration may be different on different intervals.

Worked Example: A Polynomial Denominator That Can Be Negative

Worked Example: A Quadratic With a Negative Value

Constant Multiples of the Pattern

The numerator need not equal the denominator’s derivative exactly. If it is a constant multiple, factor that constant out of the integral. For a constant \(k\),

provided \(g\) is differentiable and nonzero on the interval. The factor \(k\) stays outside the logarithm as a multiplier. This follows from the constant multiple rule for integrals; it is not generally correct to place \(k\) inside the logarithm without changing the expression.

A Reliable Recognition Process

When you suspect a logarithmic form, compare the numerator with the derivative of the denominator. A quick substitution can make the match explicit. Use this sequence:

The match must be genuine. For example, if the denominator is \(x^2+1\), its derivative is \(2x\), not \(x+1\). The expression \(\frac{x+1}{x^2+1}\) is not directly of the form \(k g'(x)/g(x)\) for that denominator with a single constant \(k\). Do not apply the logarithmic rule just because the integrand contains a fraction.

Common Mistake: Dropping the Absolute Value

A frequent incorrect step is to write \(\int \frac{g'(x)}{g(x)}\,dx=\ln(g(x))+C\) without checking whether \(g(x)>0\). That expression can fail to be real-valued when \(g(x)<0\), even though the integrand itself is defined.

For the first worked example, substituting \(x=-3\) gives a denominator of \(-29\). The incorrect proposed antiderivative \(\ln(2x^3+2x^2+7)\) would require \(\ln(-29)\), which is not defined as a real logarithm. The correction is

which is real at \(x=-3\), since \(|-29|=29>0\). Differentiating this corrected expression still returns the original quotient on any interval where the denominator is nonzero.