#\int5/((x^2+2x+2)(x-1))dx#?
#\color(crimson)\bb(\text(please help me work out the next step,))# #\color(maroon)\bb(\text(not start from the beginning))#
(I mean, I just don't want to make you spend time on the whole problem. But it's ok to do all of the work)
Partial fractions:
#5/((x^2+2x+2)(x-1))=(Ax+B)/((x^2+2x+2)(x-1))+C/(x-1)#
#A = -1, B = -3, C = 1#
#\therefore\int((-x+(-3))/(x^2+2x+2)+1/(x-1))dx#
Separating integrals:
#\int(x+(-3))/(x^2+2x+2)dx+\int1/(x-1)dx#
#\color(steelblue)(-\int(x+3)/(x^2+2x+2)dx)+\int1/(x-1)dx#
#\color(olive)\bb\text(How would I solve the blue part?)#
(I mean, I just don't want to make you spend time on the whole problem. But it's ok to do all of the work)
Partial fractions:
#5/((x^2+2x+2)(x-1))=(Ax+B)/((x^2+2x+2)(x-1))+C/(x-1)# #A = -1, B = -3, C = 1# #\therefore\int((-x+(-3))/(x^2+2x+2)+1/(x-1))dx#
Separating integrals:
#\int(x+(-3))/(x^2+2x+2)dx+\int1/(x-1)dx# #\color(steelblue)(-\int(x+3)/(x^2+2x+2)dx)+\int1/(x-1)dx#
2 Answers
Explanation:
So, we have the integral
The first problem seems to be the denominator -- this ruins the possibility of integrating as a logarithm or as an arctangent.
Let's complete the square:
Add
Thus, the completed square is
Now, we have
We can split this up:
We're going to apply the substitution
Thus, we obtain
Again, split up the first integral:
These are all simple integrals:
(No need for absolute value bars,
Simplifying, we get
We need to rewrite in terms of
So, as you can see, these can get very messy sometimes. It all comes down to using relevant substitutions and splitting up across sums and differences until you get only simple integrals.
Explanation:
Seprating integrals: