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Cilian's question at Yahoo! Answers regarding integration by partial fractions

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Chris L T521

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Jan 26, 2012
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Chris L T521

Well-known member
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Jan 26, 2012
995
Hello Cilian,

We're going to use partial fractions to evaluate this integral. With that said, we note that the partial fraction decomposition will take on the form

\[\frac{19x^2-x+4}{x(1+4x^2)} = \frac{A}{x}+\frac{Bx+C}{1+4x^2}\]

Multiplying both sides by the common denominator yields

\[19x^2-x+4 = A(1+4x^2) + (Bx+C)x\]

We now simplify the right hand side and group like terms to get

\[19x^2-x+4 = (4A+B)x^2+Cx+A\]

If we now compare the coefficients of both sides, we have the following system of equations:

\[\left\{\begin{aligned} 4A + B &= 19 \\ C &= -1 \\ A &= 4 \end{aligned}\right.\]

Luckily for us, we already have two of the solutions, so it follows now that $4(4)+B = 19 \implies B=3$. Therefore, we now see that

\[\frac{19x^2-x+4}{x(1+4x^2)} = \frac{4}{x} + \frac{3x-1}{1+4x^2}\]

Hence, we now see that

\[\int \frac{19x^2-x+4}{x(1+4x^2)}\,dx = \int\frac{4}{x}\,dx + \int\frac{3x-1}{1+4x^2}\,dx = \color{red}{\int\frac{4}{x}\,dx} + \color{blue}{\int\frac{3x}{1+4x^2}\,dx} - \color{green}{\int\frac{1}{1+4x^2}\,dx}\]

The first integral is rather straightforward; you should see that

\[\int\frac{4}{x}\,dx = \color{red}{4\ln|x|+C}\]

Next, for the second integral, we make a substitution: $u=1+4x^2\implies \,du =8x\,dx \implies \dfrac{du}{8}=x\,dx$. Thus,

\[\int\frac{3x}{1+4x^2}\,dx = \frac{3}{8}\int\frac{1}{u}\,du = \frac{3}{8}\ln|u|+C = \color{blue}{\frac{3}{8}\ln(1+4x^2)+C}\]

(Note here that we can drop absolute values since $1+4x^2>0$ for any $x$.)

For the last integral, we need to note that

\[\int\frac{1}{1+4x^2}\,dx = \int\frac{1}{1+(2x)^2}\,dx\]

To integrate, we make the substitution $u=2x\implies \,du = 2\,dx \implies \dfrac{du}{2}=\,dx$. Thus,

\[\int\frac{1}{1+4x^2}\,dx = \frac{1}{2}\int\frac{1}{1+u^2}\,du = \frac{1}{2}\arctan(u)+C = \color{green}{\frac{1}{2}\arctan(2x)+C}\]

Therefore, putting everything together, we have that

\[\int\frac{19x^2-x+4}{x(1+4x^2)}\,dx = 4\ln|x| + \frac{3}{8}\ln(1+4x^2) - \frac{1}{2}\arctan(2x) + C\]

I hope this makes sense!