Question

In Exercises \(5-14,\) evaluate the integral. $$\int \frac{2 x+16}{x^{2}+x-6} d x$$

Short answer

The integral of the given function is \(\log|x^{2}+x-6| + 15x + C\).

Step-by-step solution

Identify the derivative of the denominator

On examining the function, the denominator is \(x^{2}+x-6\). Taking the derivative of this expression, we obtain \(2x + 1\). The numerator is \(2x + 16\), which isn't exactly the derivative of the denominator but if we rewrite \((2x+16)\) as \((2x+1+15)\), we can see that \(2x+1\) is the derivative of \(x^{2}+x-6\) and \(15\) is an additional constant.

Apply the rule of integration

The function under the integral is now written in the form of \(f'(x)/f(x) + constant\). Considering only the \(\int (2x+1)/(x^{2}+x-6) dx\) part of the integral, by using the rule of integration, \(\int f'(x)/f(x) dx = \log|f(x)|\), we get \(\log|x^{2}+x-6|\).

Integrate the constant

Since \(\int du = u + C\), the integral of the constant 15 with respect to x would simply be \(15x\).

Combine the results and add the integration constant C

Since the original problem is the integral of a sum of two terms, the integral of the whole is just the sum of the integrals of the elements. Combining results from Step 2 and Step 3 and adding the constant of integration \(C\), the integral of the given function is \(\log|x^{2}+x-6| + 15x + C\).