Subject: Calculus

# Quadratic Integral

So far, we had discuss about integration by parts, inverse chain rule and substitution rule as tool in solving integrals. Such tools provide us with better techniques in solving integrals. Another technique in integration that I would like to share with you in this article is about quadratic integral.

Quadratic integral is very useful in solving integration where the numerator of the integrand is in a quadratic form which is expressed below.

\int \frac{dx}{ax^{2}+bx+c}

Obviously, the above the integral cannot be solved directly using any of the above mentioned integration techniques. That's why in this article, let us find ways to find solution of the above integral through quadratic integral.

Quadratic integral works by completing the square in the integrand's numerator, thus

Letting u=x+\frac{b}{2a}, then du=dx. Also, \frac{c}{a}+\frac{b^2}{4a^2}=\frac{4ac-b^2}{4a^2}. Thus,

would become

The second part of the integral which does not include the variable of integration is constant, thus we can assign a single variable to it. Let \frac{c}{a}+\frac{b^2}{4a^2}=A^2, then the integral would become

Now these new expression of our integral can be evaluated with two cases; value of our determinant. Determinant is the expression r=b^2-4ac. It can be seen that the value of A^2 depends on the value of the determinant r.

## Case1 (determinant r is positive)

When the determinant r is positive (b^2-4ac>0), it can be seen that A^2 is also positive, then the integral is still

which can be solved by following the steps below;

Letting s=\frac{u}{A} such that ds=\frac{du}{A}, we then get

This form of integral is easy to evaluate using trigonometric substitution, which yields

Putting back s=\frac{u}{A} , we obtain

And also inserting back u=x+\frac{b}{2a} and \frac{c}{a}+\frac{b^2}{4a^2}=A^2, we finally get

\int \frac{dx}{ax^{2}+bx+c}=\frac{1}{a\sqrt{\frac{c}{a}+\frac{b^2}{4a^2}}}\tan^{-1}{\frac{\left (x+\frac{b}{2a}\right )}{\sqrt{\frac{c}{a}+\frac{b^2}{4a^2}}}}+C

## Case 2 (determinant r is negative)

When the determinant r is negative (b^2-4ac<0), it can be seen that A^2 is also negative, then the integral would take a new form which is

The above form of integral can be solved by following the steps below;

First, let us factor the integrand in the denominator,

Then using our technique in partial fractions in integration, the term

which then takes us to

We know that

Thus,

or

Inserting back u=x+\frac{b}{2a} and \frac{c}{a}+\frac{b^2}{4a^2}=A^2, we finally get

Thus, when the discriminant r of the integral is negative, the integral is

\int \frac{dx}{ax^{2}+bx+c}=\frac{1}{2a\sqrt{\frac{c}{a}+\frac{b^2}{4a^2}}}\ln{ \frac{x+\frac{b}{2a} - A}{ x+\frac{b}{2a} + A}}

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