# #\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: