Subject: Calculus

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

ax^2+bx+c=a\left [\left (x+\frac{b}{2a}\right )^2+\left (\frac{c}{a}+\frac{b^2}{4a^2}\right )\right ]

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,

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

would become

\int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int \frac{du}{\left [\left (u\right )^2+\left (\frac{c}{a}+\frac{b^2}{4a^2}\right )\right ]}

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

\int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int \frac{du}{\left [\left (u\right )^2+\left (A^2\right )\right ]}

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

\int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int \frac{du}{\left [\left (u\right )^2+\left (A^2\right )\right ]}

which can be solved by following the steps below;

\frac{1}{a}\int \frac{du}{\left [\left (u\right )^2+\left (A^2\right )\right ]}
=\frac{1}{aA}\int \frac{du/A}{\left (u/A\right )^2+1}

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

=\frac{1}{aA}\int \frac{ds}{s^2+1}

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

=\frac{1}{aA}\tan^{-1}{s}+C

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

=\frac{1}{aA}\tan^{-1}{\frac{u}{A}}+C

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

\int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int \frac{du}{\left [\left (u\right )^2 - \left (A^2\right )\right ]}

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

First, let us factor the integrand in the denominator,

=\frac{1}{a}\int \frac{du}{\left [\left (u+A\right ) \left (u-A\right )\right ]}

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

\frac{1}{\left (u+A\right ) \left (u-A\right )}=\frac{1}{2A}\left ( \frac{1}{u-A}-\frac{1}{u+A} \right )

which then takes us to

=\frac{1}{a}\int \frac{du}{2A}\left ( \frac{1}{u-A}-\frac{1}{u+A} \right )
=\frac{1}{2aA}\int du\left ( \frac{1}{u-A}-\frac{1}{u+A} \right )
=\frac{1}{2aA}\left [\int \frac{du}{u-A}-\int \frac{du}{u+A} \right ]

We know that

\int \frac{du}{u \pm A}=\ln{ |u \pm A|}+C

Thus,

\frac{1}{2aA}\left [\int \frac{du}{u-A}-\int \frac{du}{u+A} \right ] =\frac{1}{2aA} \left [ \ln{ |u - A|} - \ln{ |u + A|}\right ]

or

\frac{1}{2aA} \ln{\frac{|u - A|}{ |u + A|}}

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

\frac{1}{2a\sqrt{\frac{c}{a}+\frac{b^2}{4a^2}}}\ln{ \frac{x+\frac{b}{2a} - A}{ x+\frac{b}{2a} + A}}

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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