How do you integrate #f(x) = (x+6)/(x+1)#?

Question

Christopher Allers · Accepted Answer

we can rewrite the function into a form which is easily integrated Think of the function like this #(x+6)/((1)(x+1))=A/1+B/(x+1) # We are doing a partial fraction decomposition.
Multiply the expression above by #(1)(x+1)# #x+6=A(x+1)+B(1) # #x+6=Ax +A +B # Equating coefficients we get #A=1 # and #A+B=6 # Since #A=1# we can conclude that #B=5# Therefore  we can rewrite as follows #int(x+6)/(x+1)dx=int1+5/(x+1)dx # Integrating we get #x+5ln|x+1|+C #

Christopher Allers · Answer

undefined To integrate the function $ f(x) = \frac{x+6}{x+1} $, you can perform polynomial long division to rewrite it as a sum of a polynomial and a proper rational function. Then, you can integrate each part separately. The steps are as follows:

1. Perform polynomial long division to rewrite $ f(x) $ as $ 1 + \frac{5}{x+1} $.
2. Integrate $ 1 $ with respect to $ x $, which gives $ x $.
3. Integrate $ \frac{5}{x+1} $ with respect to $ x $, which gives $ 5 \ln|x+1| + C $, where $ C $ is the constant of integration.

Therefore, the integral of $ f(x) $ is $ x + 5 \ln|x+1| + C $, where $ C $ is the constant of integration.