Difficult Problems—Need Practice

Partial Fraction Decomposition

1) $$\int \frac{x+20}{x^2+26x+173} dx$$

Step 1): Complete the Square (Denominator)
The denominator is an irreducible quadratic expression—it cannot be factored.

Therefore, to make it more "pliable," complete the square:

$$Ax^2+Bx+C \space = \space Ax^2+Bx+\left(\frac{B}{2}\right)^2+C-\left(\frac{B}{2}\right)^2 \\ \\$$$$x^2+26x+173 \space = \space x^2+26x+\left(\frac{26}{2}\right)^2+173-\left(\frac{26}{2}\right)^2 \\ \\$$$$x^2+26x+173 \space = \space (x^2+26x+169)+173-169 \\ \\ $$$$x^2+26x+173 \space = \space (x+13)^2+4 \\ \\ $$

Step 2): Split the integrand into multiple integrals and integrate using appropriate techniques
Completing the square will help in a later step, but it made sense to get it out of the way, first.

Now, split the integrand by splitting up the rational function:

$$\int \frac{x+20}{x^2+26x+173} dx \space = \space \int \frac{x+13}{x^2+26x+173} dx \quad + \quad \int \frac{7}{x^2+26x+173}dx$$

Step 3): Integrate, one integral at a time
The first integral can be integrated using the Substitution Rule.

$$ \text{Integral 1:} \quad \int \frac{x+13}{x^2+26x+173} dx \space = \space \frac{1}{2} \int \frac{2(x+13)}{x^2+26x+173} dx \\ \\ = \frac{1}{2} \int \frac{2x+26}{x^2+26x+173} dx \\ \\ = \frac{1}{2} \ln(x^2+26x+173) + C \quad \tiny\text{Note: normally, the argument to the} \space \textbf{natural log function} \space \text{must be placed within an} \space \textbf{absolute value sign,} \\ \tiny\text{but} \space x^2+26x+173 \space \text{cannot be negative, so the absolute value sign is unnecessary.} $$$$ \\ \text{Integral 2:} \quad \int \frac{7}{x^2+26x+173} dx \space = \space 7\int \frac{1}{x^2+26x+173} dx \\ \\ = 7 \int \frac{1}{(x+13)^2+4} dx $$$$ \tiny\text{The most appropriate integration technique for this integral will be} \space \textbf{Trigonometric Substitution.} \\ \implies \text{Let} \space u = 2\tan\theta \\ \implies \quad du = 2\sec^2\theta d\theta \\ \implies \quad \tan\theta = \frac{u}{2} \\ \implies \quad \theta = \arctan\left(\frac{u}{2}\right) $$$$ \implies \quad 7 \int \frac{1}{(x+13)^2+4} dx \quad = \quad 7 \int \frac{2\sec^2\theta}{(2\tan\theta)^2+4} d\theta \\ \\ \implies 7 \int \frac{2\sec^2\theta}{4\tan^2\theta+4} d\theta \quad = \quad 7 \int \frac{2\sec^2\theta}{4(\tan^2\theta+1} d\theta \\ \\ \implies 7 \int \frac{2\sec^2\theta}{4\sec^2\theta} d\theta \\ \\ \implies 7 \int \frac{1}{2} d\theta \\ \\ \implies \frac{7}{2}\theta + C \quad \tiny\text{Now, do the back substitution using the value for} \space \theta \space \tiny\text{and for} \space u \space \tiny\text{found previously.} \\ \\ \theta = \arctan\left(\frac{u}{2}\right) \quad \implies \quad \frac{7}{2}\theta + C \space = \space \frac{7}{2}\arctan\left(\frac{x+13}{2}\right) + C $$

Step 4): Combine
Combine the two antiderivatives to form the final solution. $$ \\ \int \frac{x+20}{x^2+26x+173} dx = \frac{1}{2}\ln(x^2+26x+173) + \frac{7}{2}\arctan\left(\frac{x+13}{2}\right) + C \\ \\ \small \text{Done!} $$



2) $$\int \frac{x^3+36}{x^2+36} dx$$