Euler Substitution with Positive Constant Term

Why Use a Linear Relation with the Radical?

A square root such as \(\sqrt{ax^2+bx+c}\) makes an integrand algebraic rather than rational. When \(c>0\), the quadratic has the known point \((0,\sqrt c)\) on its associated curve \(y^2=ax^2+bx+c\). A line through that point can be used to express \(x\) and \(y\) rationally in terms of the line's slope.

This is the Euler substitution for a quadratic radical with positive constant term. Its usefulness is concrete: after the substitution, expressions built rationally from \(x\) and \(y\), together with \(dx\), become rational functions of a new parameter. They can then be handled using rational integration techniques.

Setup: Let \(a,b\in\mathbb R\), \(c>0\), and let \(y=\sqrt{ax^2+bx+c}\) denote the nonnegative square root. Set \(s=\sqrt c\) and introduce the parameter \(t\) by \(y=tx+s\).

The formulas below apply wherever \(a-t^2\ne0\). The displayed formula for \(y\) must also be nonnegative to agree with the original square root. Strict positivity is not required for the formulas for \(x\) and \(y\); it is required only in steps that divide by \(y\).

Deriving the Rational Formulas

Substitute \(y=tx+s\) into \(y^2=ax^2+bx+c\), and use \(s^2=c\):

$$ (tx+s)^2=ax^2+bx+s^2. $$

After cancelling \(s^2\) and collecting terms, the relation becomes

x\bigl((t^2-a)x+2st-b\bigr)=0.

When \(x\ne0\), the second factor must vanish. Solving it for \(x\), and then using \(y=tx+s\), gives the rational expressions below. The resulting expressions also parametrize points with \(x=0\) when their numerator gives \(x=0\); those points can be checked directly in the original relation.

x=\frac{2st-b}{a-t^2}, \qquad y=\frac{s(a+t^2)-bt}{a-t^2}.

To find the differential, differentiate the formula for \(x\) with respect to \(t\). The quotient rule gives

\frac{dx}{dt} = \frac{2s(a-t^2)+2t(2st-b)}{(a-t^2)^2} = \frac{2\bigl(s(a+t^2)-bt\bigr)}{(a-t^2)^2} = \frac{2y}{a-t^2}.

Thus the substitution's differential is

dx=\frac{2y}{a-t^2}\,dt.

Where \(y\ne0\), division by \(y\) gives an especially useful cancellation:

\frac{dx}{y}=\frac{2}{a-t^2}\,dt.

The differential formula itself remains valid when \(y=0\): in that case \(dx/dt=0\), provided \(a-t^2\ne0\). The expression \(dx/y\), however, cannot be obtained by dividing by \(y\) at such a point.

Why the Substitution Rationalizes

The derivation explains why the formulas work, but it is also useful to verify that the parametrized expressions satisfy the original quadratic equation. From the definition of \(x\),

$$ (a-t^2)x=2st-b. $$

Since \(y=tx+s\), the difference between \(y^2\) and \(ax^2+bx+s^2\) is

\begin{aligned} y^2-(ax^2+bx+s^2) &=(tx+s)^2-(ax^2+bx+s^2)\\ &=x\bigl((t^2-a)x+2st-b\bigr)\\ &=0. \end{aligned}

So the parametrized \(x\) and \(y\) lie on the required quadratic curve. If the displayed \(y\) is nonnegative, it is the original nonnegative square root. Both \(x\) and \(y\) are rational functions of \(t\), and \(dx/dt\) is rational as well. Consequently, a rational expression in \(x\) and \(y\), multiplied by \(dx\), becomes a rational expression in \(t\), wherever the denominators involved are nonzero.

Local use: For an integral, work on an interval where the parameterization has the required square-root branch, \(a-t^2\ne0\), and the transformed integrand is defined. If using \(dx/y\), also require \(y\ne0\) on the interval. These conditions allow the substitution to be used as an ordinary change of variable.

Worked Example: Integrating the Reciprocal of \(\sqrt{4x^2+1}\)

Worked Example: A Reciprocal Radical

Evaluate \(\int \frac{dx}{\sqrt{4x^2+1}}\). Here \(a=4\), \(b=0\), \(c=1\), and \(s=1\). The Euler formulas are

x=\frac{2t}{4-t^2}, \qquad y=\frac{t^2+4}{4-t^2}, \qquad dx=\frac{2y}{4-t^2}\,dt.

The branch \(y>0\) is obtained for \(-2<t<2\). In that interval, \(4-t^2>0\) and \(t^2+4>0\). Since \(y\ne0\), substitute into the integral:

\int\frac{dx}{y} = \int\frac{2}{4-t^2}\,dt = \frac12\ln\left|\frac{2+t}{2-t}\right|+C.

The parameter can be expressed in terms of \(x\) and \(y\). From \(y=tx+1\), \(t=(y-1)/x\) when \(x\ne0\). Rationalizing that expression gives \(t=4x/(y+1)\), which is also defined at \(x=0\). Thus an antiderivative in the original variable is

\frac12\ln\left|\frac{2+\frac{4x}{y+1}}{2-\frac{4x}{y+1}}\right|+C, \qquad y=\sqrt{4x^2+1}.

To check, differentiate the antiderivative first with respect to \(t\). Its derivative is \(2/(4-t^2)\), while \(dx/dt=2y/(4-t^2)\). Therefore its derivative with respect to \(x\) is \(\bigl(2/(4-t^2)\bigr)\big/\bigl(2y/(4-t^2)\bigr)=1/y\), as required.

Worked Example: A Radical in Both the Denominator and the Integrand

Worked Example: Integrating \(1/(xy)\)

Consider \(\int \frac{dx}{x\sqrt{x^2+3x+2}}\) on an interval where \(x>0\) and the radical is positive. Set \(y=\sqrt{x^2+3x+2}\), so \(a=1\), \(b=3\), \(c=2\), and \(s=\sqrt2\). The parameter formulas are

x=\frac{2\sqrt2\,t-3}{1-t^2}, \qquad y=\frac{\sqrt2(1+t^2)-3t}{1-t^2}, \qquad \frac{dx}{y}=\frac{2}{1-t^2}\,dt.

Because \(x>0\), \(1/x=(1-t^2)/(2\sqrt2\,t-3)\). Hence

\int\frac{dx}{xy} = \int \frac{1-t^2}{2\sqrt2\,t-3} \frac{2}{1-t^2}\,dt = \int\frac{2}{2\sqrt2\,t-3}\,dt = \frac{1}{\sqrt2}\ln|2\sqrt2\,t-3|+C.

On the stated interval, \(t=(y-\sqrt2)/x\). This gives the antiderivative

\frac{1}{\sqrt2} \ln\left|2\sqrt2\,\frac{y-\sqrt2}{x}-3\right|+C, \qquad y=\sqrt{x^2+3x+2}.

For a derivative check, the parameter equation \(x=(2\sqrt2\,t-3)/(1-t^2)\) implies \(t'=(1-t^2)/(2y)\), by the differential formula. Differentiating the proposed antiderivative with respect to \(x\) gives

\frac{1}{\sqrt2} \frac{2\sqrt2\,t'}{2\sqrt2\,t-3} = \frac{1-t^2}{y(2\sqrt2\,t-3)} = \frac{1}{xy}.

The last equality follows from the displayed formula for \(x\), so the derivative matches the integrand.

Worked Example: A Definite Integral with Exact Parameter Limits

Worked Example: Evaluating a Definite Integral

Evaluate \(\int_0^{3/4}\frac{dx}{\sqrt{x^2+1}}\). Here \(a=1\), \(b=0\), \(c=1\), and \(s=1\). The substitution gives

x=\frac{2t}{1-t^2}, \qquad y=\frac{1+t^2}{1-t^2}, \qquad \frac{dx}{y}=\frac{2}{1-t^2}\,dt.

At \(x=0\), \(y=1\) and \(t=(y-1)/x\) is not directly defined. The parameter relation \(y=tx+1\) and the formula for \(x\) show that this endpoint corresponds to \(t=0\). At \(x=3/4\), \(y=\sqrt{(3/4)^2+1}=5/4\), so \(t=(y-1)/x=(1/4)/(3/4)=1/3\). The transformed integral is therefore

\int_0^{3/4}\frac{dx}{\sqrt{x^2+1}} = \int_0^{1/3}\frac{2}{1-t^2}\,dt = \left[\ln\left|\frac{1+t}{1-t}\right|\right]_0^{1/3} = \ln 2.

The endpoint values are consistent with the parametrization: at \(t=0\), \(x=0\) and \(y=1\); at \(t=1/3\), \(x=(2/3)/(8/9)=3/4\) and \(y=(10/9)/(8/9)=5/4\). Differentiating the corresponding antiderivative with respect to \(x\) gives \(1/y\), as in the first worked example.

A Sign Error to Avoid

The sign in the formula for \(x\) is easy to reverse. For instance, replacing \(2st-b\) with \(b-2st\) would produce the wrong parametrization. Take \(a=4\), \(b=0\), \(c=1\), and \(t=1/2\). The correct formula gives

x=\frac{1}{4-\frac14}=\frac{4}{15}, \qquad y=tx+1=\frac{17}{15}.

Indeed, \(4x^2+1=4(16/225)+1=289/225=y^2\). The incorrect reversed sign would give \(x=-4/15\) and \(y=13/15\), for which \(y^2=169/225\), while \(4x^2+1=289/225\). They do not agree. The reliable derivation is to expand \(y=tx+s\), cancel \(s^2=c\), and solve the resulting equation with its signs unchanged.

Using the Method Reliably

The substitution is a method for changing the variables, not a guarantee that every resulting integral is immediately simple. Once the transformed expression is rational, its denominator may still have factors or poles that require attention. The interval of integration must not cross a point where the original integrand is undefined or where the chosen parameterization fails.

Keep the branch condition visible throughout the calculation. The rational formula for \(y\) represents the nonnegative square root only where it is nonnegative. It may equal zero; that does not invalidate the formulas for \(x\) and \(y\) if \(a-t^2\ne0\). But whenever the calculation uses \(dx/y\), it must be restricted to points where \(y\ne0\), or treated by an appropriate limiting argument at an endpoint.

The central advantage is that the relation \(y=tx+\sqrt c\) uses the known point at \(x=0\) to turn the quadratic equation into rational formulas. Differentiating those formulas supplies the differential, and the radical can then be handled through ordinary rational integration.

Check Your Understanding

Use the formulas and conditions from this tutorial to answer the following questions.

Starting from \(y=tx+\sqrt c\), what equation leads to the formula for \(x\)?
For what values of \(t\) is the displayed formula for \(x\) undefined?
Why must the displayed formula for \(y\) be nonnegative, and is strict positivity always necessary?
Derive \(dx/y=2\,dt/(a-t^2)\), and state the condition needed for this division.
For \(a=1\), \(b=0\), and \(c=1\), find the parameter value corresponding to \(x=3/4\).

Euler Substitution with Positive Constant Term

What You'll Learn

Why Use a Linear Relation with the Radical?

Deriving the Rational Formulas

Why the Substitution Rationalizes