A Tricky Integral

$$\int \frac{3x+1}{x(x^2+4)}dx$$

This integral requires the use of the following techniques:

• Partial fractions
• Integration by substitution
• Algebraic manipulation
• Antiderivatives involving natural logarithms
• Antiderivatives involving arctangent

Solution

Taking note that this integral is of the form

$$\int \frac{\text{polynomial of lower degree}}{\text{factored polynomial}}dx,$$

we are motivated to perform a partial fraction decomposition on the integrand. Let $A,$ $B,$ and $C$ be constants. For the correct choice of constants, the following relationship should be true for all values of $x.$

$$\frac{3x+1}{x(x^2+4)}=\frac{A}{x}+\frac{Bx+C}{x^2+4}$$

Multiply both sides by $x(x^2+4)$ to clear the denominators.

$$3x+1=A(x^2+4)+B{x}^2+Cx$$

Plugging in $x=0$ readily yields $A=\frac{1}{4}.$

$$\begin{array}{c}3x+1=\frac{1}{4}(x^2+4)+B{x}^2+Cx\\ -\frac{1}{4}{x}^2+3x=B{x}^2+Cx\end{array}$$

Plugging in $x=\pm 1$ yields the following system of equations.

$$\{\begin{array}{ll}\frac{11}{4}=B+C& \text{for x=1}\\ -\frac{13}{4}=B-C& \text{for x=−1}\end{array}$$

Adding these two equations gives $-\frac{1}{2}=2B,$ so $B=-\frac{1}{4}.$ Back-substituting this into either equation above gives $C=3$. This completes the partial fraction decomposition. Split up the main integral into three pieces according to this decomposition.

$$\begin{array}{rl}\int \frac{3x+1}{x(x^2+4)}dx& =\int \frac{\frac{1}{4}}{x}+\frac{-\frac{1}{4}x+3}{x^2+4}dx\\ & =\frac{1}{4}\underset{\text{integral A}}{\underbrace{\int \frac{1}{x}dx}}-\frac{1}{4}\underset{\text{integral B}}{\underbrace{\int \frac{x}{x^2+4}dx}}+3\underset{\text{integral C}}{\underbrace{\int \frac{1}{x^2+4}dx}}\end{array}$$

Integral A: This integral is straightforward.

$$\int \frac{1}{x}dx=\ln |x|+C$$

Integral B: This integral can be computed by letting $u=x^2+4,$ so $\frac{1}{2}du=xdx.$

$$\begin{array}{rl}\int \frac{x}{x^2+4}dx& =\frac{1}{2}\int \frac{1}{u}du\\ & =\frac{1}{2}\ln |u|+C\\ & =\frac{1}{2}\ln |x^2+4|+C\\ & =\frac{1}{2}\ln (x^2+4)+C\end{array}$$

I dropped the absolute value signs in the last step since $x^2+4$ is always positive.

Integral C: This integral can be manipulated into a form of an arctangent derivative and integrated by ultimately letting $u=\frac{x}{2},$ which gives $2du=dx.$

$$\begin{array}{rl}\int \frac{1}{x^2+4}dx& =\frac{1}{4}\int \frac{1}{(\frac{x}{2}{)}^2+1}dx\\ & =\frac{1}{2}\int \frac{1}{u^2+1}du\\ & =\frac{1}{2}\arctan u+C\\ & =\frac{1}{2}\arctan \left(\frac{x}{2}\right)+C\end{array}$$

Putting it all together: The results of these three individual integrations can be put together to provide the final answer.

$$\begin{array}{rl}\int \frac{3x+1}{x(x^2+4)}dx& =\frac{1}{4}\underset{\text{integral A}}{\underbrace{\int \frac{1}{x}dx}}-\frac{1}{4}\underset{\text{integral B}}{\underbrace{\int \frac{x}{x^2+4}dx}}+3\underset{\text{integral C}}{\underbrace{\int \frac{1}{x^2+4}dx}}\\ & =\frac{1}{4}\ln |x|-\frac{1}{4}\cdot \frac{1}{2}\ln |x^2+4|+3\cdot \frac{1}{2}\arctan \left(\frac{x}{2}\right)+C\\ & =\boxed{\frac{1}{4}\ln |x|-\frac{1}{8}\ln (x^2+4)+\frac{3}{2}\arctan \left(\frac{x}{2}\right)+C}\end{array}$$

Optionally, the natural logarithms can be combined together using properties of logarithms.

$$\int \frac{3x+1}{x(x^2+4)}dx=\boxed{\frac{1}{4}\ln \left(\frac{|x|}{\sqrt{x^2+4}}\right)+\frac{3}{2}\arctan \left(\frac{x}{2}\right)+C}$$