How can I evaluate this integral, if the denominator has a quadratic factors e.i b2−4ac&lt;0∫xdxx2+6x+13 by partial decomposition?

∫xdxx2+6x+13=12∫(2x+6)dxx2+6x+13−∫3dxx2+6x+13The first term in the integral above has numerator of integrand as derivative of its denominator. So integral can be evaluated by substituting denominator as a variable.We now need to find:∫3dxx2+6x+13The general trick is to use completing squares if the quadratic equation has no real roots. So we can write the integral as:∫3dx(x+3)2+22Now substitute x+3=t and then you get:∫3dtt2+22Now complete this using standard integral of inverse tangent function.

Resolved: "How can I evaluate this integral, if the..."

High School
Algebra
Algebra I
Quadratic function and equation

Rowan Callahan

Answered question

2022-03-27

How can I evaluate this integral, if the denominator has a quadratic factors e.i

b^{2} - 4 a c < 0

\int \frac{x d x}{x^{2} + 6 x + 13}

by partial decomposition?

Answer & Explanation

glikozyd3s68

Beginner2022-03-28Added 16 answers

\int \frac{x d x}{x^{2} + 6 x + 13}

= \frac{1}{2} \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 13} - \int \frac{3 d x}{x^{2} + 6 x + 13}

The first term in the integral above has numerator of integrand as derivative of its denominator. So integral can be evaluated by substituting denominator as a variable.
We now need to find:

\int \frac{3 d x}{x^{2} + 6 x + 13}

The general trick is to use completing squares if the quadratic equation has no real roots. So we can write the integral as:

\int \frac{3 d x}{{(x + 3)}^{2} + 2^{2}}

Now substitute

x + 3 = t

and then you get:

\int \frac{3 d t}{t^{2} + 2^{2}}

Now complete this using standard integral of inverse tangent function.

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