Why split the numerator?
An integral with a linear numerator and a quadratic square root can look more complicated than it is. The key is that differentiating the quadratic radicand produces a linear expression. Part of the numerator can therefore be integrated directly by the chain rule; the remaining constant is handled by a reciprocal-square-root formula.
This method is useful precisely because it separates two tasks: recognizing a derivative already present in the integrand, and integrating the reciprocal square root of the same quadratic. The quadratic’s type determines which reciprocal-square-root formula applies, and its positive domain determines where the antiderivative is valid.
The numerator decomposition
We seek constants \(\alpha\) and \(\beta\) such that \(mx+n=\alpha(2ax+b)+\beta\). Matching the coefficients of \(x\) and the constant terms gives \(\alpha=\frac{m}{2a}\) and \(\beta=n-\frac{mb}{2a}\). Thus the integrand separates into a derivative term and a constant multiple of \(1/\sqrt{Q(x)}\).
The first term has a direct antiderivative, because \(Q(x)>0\) on the interval under consideration:
For the second term, use an antiderivative of \(1/\sqrt{Q(x)}\) suited to the quadratic. To state the necessary formulas, write the discriminant as \(\Delta=b^2-4ac\). Each formula below applies on an interval where \(Q(x)>0\).
When \(a>0\) and \(\Delta=0\), the radicand is positive on both intervals excluding \(r\), and \(\sqrt{Q(x)}=\sqrt a\,|x-r|\). Therefore, an antiderivative of \(1/\sqrt{Q(x)}\) is \(\frac{1}{\sqrt a}\ln(x-r)\) on \(x>r\), and \(-\frac{1}{\sqrt a}\ln(r-x)\) on \(x<r\). If \(a<0\), there is an interval where \(Q>0\) only when \(\Delta>0\); if \(a<0\) and \(\Delta\le 0\), there is no open interval on which this real-valued integrand is defined.
The general method and why it works
Combining the split with the appropriate reciprocal-square-root antiderivative \(J(x)\) gives the general form. Here \(J'(x)=1/\sqrt{Q(x)}\) on the interval being considered.
The expression is a local antiderivative on each interval where \(Q(x)>0\); the constant \(C\) may differ between intervals. The coefficient of the square-root term is \(m/a\) because the split gives \(m/(2a)\) times \(Q'/\sqrt Q\), whose antiderivative is \(2\sqrt Q\).
For \(Q(x)=ax^2+bx+c\), use \(Q'(x)=2ax+b\).
Write \(mx+n=\frac{m}{2a}(2ax+b)+\left(n-\frac{mb}{2a}\right)\).
The derivative part gives \(\frac{m}{a}\sqrt Q\); integrate the constant multiple of \(1/\sqrt Q\) using the formula that matches the quadratic.
This also justifies the method in full: differentiating the proposed result gives \(\frac{m}{a}\frac{Q'}{2\sqrt Q}+\left(n-\frac{mb}{2a}\right)\frac{1}{\sqrt Q}\). The coefficient of \(Q'\) is \(m/(2a)\), so recombining the numerator gives \(\frac{(m/(2a))(2ax+b)+n-mb/(2a)}{\sqrt Q}=\frac{mx+n}{\sqrt Q}\), as required.
Worked examples
Worked Example: A quadratic positive for every real \(x\)
Evaluate \(\int \frac{3x+8}{\sqrt{x^2+4x+8}}\,dx\).
Here \(a=1\), \(b=4\), \(c=8\), \(m=3\), and \(n=8\). The discriminant is \(\Delta=4^2-4(1)(8)=-16\), so the radicand is positive for every real \(x\). The split coefficients are \(\alpha=\frac{3}{2}\) and \(\beta=8-\frac{3\cdot4}{2}=2\). Thus \(3x+8=\frac32(2x+4)+2\).
To check, set \(Q=x^2+4x+8\) and \(L=2\sqrt Q+2x+4\). Then \(L'=Q'/\sqrt Q+2\), while \(L/\sqrt Q=2+Q'/\sqrt Q\), so \(\frac{d}{dx}\ln|L|=1/\sqrt Q\). Differentiating the proposed answer therefore gives \(\frac{3Q'}{2\sqrt Q}+\frac{2}{\sqrt Q}=\frac{3(x+2)+2}{\sqrt Q}=\frac{3x+8}{\sqrt Q}\).
Worked Example: A radicand positive outside its roots
Evaluate \(\int \frac{2x+3}{\sqrt{x^2-4}}\,dx\).
Now \(Q=x^2-4\), so \(a=1\), \(b=0\), \(m=2\), and \(n=3\). Its discriminant is \(\Delta=16\), and \(Q>0\) on \(x>2\) and \(x<-2\). The split is \(2x+3=1\cdot(2x)+3\), so \(\alpha=1\) and \(\beta=3\). On either domain interval the logarithmic formula applies:
For the check, let \(L=2\sqrt{x^2-4}+2x\). Then \(L'=2x/\sqrt{x^2-4}+2=L/\sqrt{x^2-4}\). Consequently, the derivative of the logarithm is \(1/\sqrt{x^2-4}\). Differentiating the full answer gives \(2x/\sqrt{x^2-4}+3/\sqrt{x^2-4}=(2x+3)/\sqrt{x^2-4}\). The formula holds separately on each interval; the absolute value keeps the logarithm defined when \(L<0\).
Worked Example: A bounded interval of definition
Evaluate \(\int \frac{4x+5}{\sqrt{9-x^2}}\,dx\).
Here \(Q=9-x^2\), with \(a=-1\), \(b=0\), \(m=4\), and \(n=5\). The discriminant is \(\Delta=36>0\), and \(Q>0\) on \(-3<x<3\). The split coefficients are \(\alpha=4/(2(-1))=-2\) and \(\beta=5\), giving \(4x+5=-2(-2x)+5\). The reciprocal-square-root antiderivative is \(\arcsin(x/3)\), equivalently the \(a<0\) formula above. Therefore:
For verification, the derivative of \(-4\sqrt{9-x^2}\) is \(4x/\sqrt{9-x^2}\). The derivative of \(5\arcsin(x/3)\) is \(\frac{5}{\sqrt{9-x^2}}\), since \(\frac{d}{dx}\arcsin(x/3)=1/\sqrt{9-x^2}\) on this interval. Their sum is the original integrand.
Worked Example: A repeated-root radicand
Evaluate \(\int \frac{3x+1}{\sqrt{(x-2)^2}}\,dx\), separately on \(x>2\) and \(x<2\).
Here \(Q=(x-2)^2\), so \(a=1\), \(b=-4\), \(m=3\), and \(n=1\). The split coefficients are \(\alpha=3/2\) and \(\beta=1-\frac{3(-4)}2=7\). In particular, \(\sqrt Q=|x-2|\), not \(x-2\) on the entire domain. On \(x>2\), an antiderivative of \(1/\sqrt Q\) is \(\ln(x-2)\). On \(x<2\), it is \(-\ln(2-x)\). The resulting answers are:
For \(x>2\), the derivative is \(3+7/(x-2)=(3x+1)/(x-2)\). For \(x<2\), it is \(-3+7/(2-x)=(3x+1)/(2-x)\). Since \(|x-2|\) equals \(x-2\) on the first interval and \(2-x\) on the second, both derivatives match the integrand on their respective domains.
A common mistake: losing the absolute value
For the repeated-root example, writing \(\sqrt{(x-2)^2}=x-2\) for every \(x\) is incorrect: the square root is \(|x-2|\). For instance, at \(x=1\), the actual denominator is \(1\), while \(x-2=-1\). Ignoring this distinction can produce an expression whose derivative is wrong on \(x<2\). The correction is to work on each interval separately, use the correct value of \(|x-2|\), and check the resulting derivative there.
A second error is to treat the entire numerator as a multiple of \(Q'\). The constant left over after the split is generally not zero. For example, in the first worked example, \(3x+8=\frac32Q'+2\), not merely \(\frac32Q'\). Omitting the \(2\) would omit the logarithmic contribution and would fail the derivative check.
Choosing and checking the result
Before integrating, confirm that \(a\ne0\), identify where \(Q>0\), and compute both split coefficients. Then select the reciprocal-square-root formula using the sign of \(a\) and the discriminant. If the discriminant is zero, keep the domain intervals separate and preserve the absolute value in \(\sqrt{a(x-r)^2}\). These checks prevent formula selection from being detached from the actual domain of the integrand.
The method matters because it turns a family of integrals into a repeatable procedure rather than a collection of unrelated tricks. Its final check is especially simple: differentiate the square-root term and the reciprocal-square-root term separately, then combine their numerators. If the result is \(mx+n\), the decomposition and the antiderivative agree.
Check Your Understanding
For each question, keep the domain of the integrand in view and check any proposed antiderivative by differentiation.
- For \(Q(x)=2x^2+6x+5\) and numerator \(4x+7\), find \(\alpha\) and \(\beta\) in \(4x+7=\alpha Q'(x)+\beta\).
- For \(Q(x)=x^2-9\), where is \(1/\sqrt{Q(x)}\) defined over the real numbers, and which reciprocal-square-root formula applies?
- For \(Q(x)=4-x^2\), state the interval on which the integrand is real and identify the inverse-trigonometric form for its reciprocal-square-root antiderivative.
- Why must \(\sqrt{(x-1)^2}\) be written as \(|x-1|\), and what is the domain split needed for integration?
- Explain why a constant term remains after splitting a general linear numerator, and describe how differentiating the final answer checks that the split was correct.
