How do I solve
Integral of 7/(16-x^2)
I know I must break down (16-x^2) into (x+4)(-x+4), but after I do that what is next?
Using the method of partial fractions, convince yourself that
7/(16-x^2)= (7/8)*[1/(4+x) + 1/(4-x)]
The two additive
terms can be integrated by the method of substitution.
4+x -> u etc. You will end up with terms that are the log of 4+x and 4-x.
To solve the integral ∫(7/(16-x^2)) dx, you correctly identified that you need to break down the denominator (16-x^2) into (x+4)(-x+4) using the difference of squares formula.
After doing that, you can proceed with the method of partial fractions to rewrite the integral as a sum of simpler fractions. In this case, you have:
∫(7/(16-x^2)) dx = ∫[(7/8)*[1/(4+x) + 1/(4-x)]] dx
Now, you can integrate each term separately. Let's focus on the term 1/(4+x) and assume you follow the same procedure for the other term:
Let u = 4 + x
Then, du = dx
Our integral becomes:
∫(7/8)*(1/u) du
This simplifies to:
(7/8) ∫(1/u) du
Now, we integrate 1/u with respect to u. The integral of 1/u is the natural logarithm (ln) of |u| plus a constant. Therefore:
(7/8) ∫(1/u) du = (7/8) ln|u| + C
Remembering that we let u = 4 + x, we substitute it back:
(7/8) ln|4 + x| + C
Similarly, you would follow the same procedure to integrate the other term 1/(4-x). Once you have integrated both terms, you can combine the results to get the final solution for the integral.