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Evaluate the integral \int \frac{x^2 + 3}{(x^2 + 9x - 4)^2} \, dx

Question:

Evaluate the integral {eq}\int \frac{x^2 + 3}{(x^2 + 9x - 4)^2} \, dx {/eq}

Integrating with Partial Fraction Decomposition

When integrating a rational function {eq}f(x) = \dfrac{p(x)}{q(x)} {/eq}, where {eq}q(x)\neq 0 {/eq}, {eq}p(x) {/eq} and {eq}q(x) {/eq} are polynomials and the degree of {eq}q(x) {/eq} is greater than the degree of {eq}p(x) {/eq}, using the technique of rewriting the integrand in its partial fraction decomposition (also called partial fraction expansion) prior to integrating is very common. The partial fraction decomposition of a rational function is found by reversing the procedure of finding a common denominator and adding. To find the partial fraction decomposition of a rational function, begin by fully factoring the denominator. For each linear factor {eq}ax+b {/eq} in the denominator, there is a corresponding term in the partial fraction decomposition of the form {eq}\dfrac{A}{ax+b} {/eq}. For every irreducible quadratic factor {eq}ax^2+bx + c {/eq} in the denominator, there is a corresponding term in the partial fraction decomposition of the form {eq}\dfrac{Ax+B}{ax^2+bx+c} {/eq}. For example, if your rational function, after factoring the denominator, looks like {eq}\dfrac{p(x)}{(ax+b)(cx^2+dx+e)^2} {/eq} the partial fraction decomposition would be of the form {eq}\dfrac{p(x)}{(ax+b)(cx^2+dx+e)^2} = \dfrac{A}{ax+b} + \dfrac{Bx+C}{cx^2+dx+e} + \dfrac{Dx+E}{(cx^2+dx+e)^2} {/eq}. To find the constants {eq}A,B,C,D,E {/eq} multiply both sides of the equation by the factored denominator and set up a system of equations by comparing coefficients on both sides. After the constants are found, rewrite the original integral using the partial fraction decomposition and integrate term by term.

Answer and Explanation:

To integrate {eq}\displaystyle\int \frac{x^2 + 3}{(x^2 + 9x - 4)^2} \, dx {/eq} , first find the partial fraction decomposition of the integrand.

The denominator is already fully factored, so we have

{eq}\dfrac{x^2 + 3}{(x^2 + 9x - 4)^2} = \dfrac{Ax+B}{x^2+9x-4} + \dfrac{Cx+D}{(x^2+9x-4)^2}\\ x^2 + 3 = (Ax+B)(x^2+9x-4) + Cx + D\\ x^2 + 3 = Ax^3 + (9A+B)x^2 + (9B+C-4A)x + D-4B {/eq}

Comparing coefficients on both sides of the equation, we have the system of equations

{eq}0=A\\ 1=9A+B\\ 0=9B+C-4A\\ 3=D-4B {/eq}

which has solution {eq}A=0 , B=1 , C=-9 , D=7 {/eq}

Rewriting our integral with the partial fraction decomposition, we have

{eq}\displaystyle\int \frac{x^2 + 3}{(x^2 + 9x - 4)^2} \, dx = \int\left(\frac{1}{x^2+9x-4} + \frac{-9x+7}{(x^2+9x - 4)^2} \right) dx {/eq}

Normally, after reaching this point, the integral can be calculated easily with some u-substitution. However, this integral is much more complicated. Using a computer algebra system, we have

{eq}= \dfrac{-2\tanh^{-1}\left(\frac{2x+9}{\sqrt{97}}\right)}{\sqrt{97}} + \dfrac{9-95x}{97(x^2 + 9x -4)} + \dfrac{190\tanh^{-1}\left(\frac{2x+9}{\sqrt{97}}\right)}{97\sqrt{97}} + c = \dfrac{-4\tanh^{-1}\left(\frac{2x+9}{\sqrt{97}}\right)}{\sqrt{97}} + \dfrac{9-95x}{97(x^2 + 9x -4)}+c {/eq}


Learn more about this topic:

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How to Integrate Functions With Partial Fractions

from Math 104: Calculus

Chapter 13 / Lesson 9
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