Replace the ? with a function of x that will make the integrand equal to the derivative of a product, and then find the
antiderivative. Choose 0 for the constant of integration.
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Question

Quizplus · Accepted Answer

We want to find a function of x that makes the integrand equal to the derivative of a product. One common way to do this is to use the product rule for differentiation, which states that (fg)' = f'g + fg'.

Looking at the integrand, we see that the expression x^2 - x - 2 is quadratic, so it makes sense to try factoring it. We have:

x^2 - x - 2 = (x - 2)(x + 1)

Now we can use the product rule with f(x) = x - 2 and g(x) = ln|x + 1|:

f'(x) = 1, g'(x) = 1/(x + 1)

f(x)g'(x) = (x - 2)/(x + 1)

f(x)g'(x) + f'(x)g(x) = (x - 2)/(x + 1) + 1/(x + 1) = (x - 1)/(x + 1)

This matches the integrand, so we can write:

∫(x^2 - x - 2)/(x + 1) dx = ∫[(x - 2)/(x + 1) + 1/(x + 1)] dx
= ∫(x - 2)/(x + 1) dx + ∫1/(x + 1) dx

The first integral is a natural log:

∫(x - 2)/(x + 1) dx = ∫(1 - 3/(x + 1)) dx = x - 3ln|x + 1|

The second integral is also a natural log:

∫1/(x+1) dx = ln|x+1|

Putting it all together, we have:

∫(x^2 - x - 2)/(x + 1) dx = (x - 3ln|x + 1|) + ln|x + 1| + C = x - 2ln|x + 1| + C

We choose C = 0 based on the instructions. The integrand is $7x\cos(2x)$, which is the derivative of $-\frac{7}{2}\sin(2x)$. Thus, we can write $\int 7x\cos(2x) dx = -\frac{7}{2}\sin(2x) + C$ where we choose $C=0$ to get the desired antiderivative.

Replace the ? with a Function of X That Will