Simple integral: #int {-3x+5}/{x^2-2x+5}dx=?#

#int (-3x+5)/(x^2-2x+5)*dx#

=#-int (3x-5)/(x^2-2x+5)*dx#

=#-int (3x-3-2)/(x^2-2x+5)*dx#

=#-int (3x-3)/(x^2-2x+5)*dx#+#int 2/(x^2-2x+5)*dx#

=#int 2/((x-1)^2+4)*dx#-#3/2int (2x-2)/(x^2-2x+5)#

=#arctan((x-1)/2)-3/2ln(x^2-2x+5)#

#int (-3x+5)/(x^2-2x+5)dx#

#=int (-3x+5-2+2)/(x^2-2x+5)dx#

#=int (-3x+3)/(x^2-2x+5)+2/(x^2-2x+5)dx#

#=-int(3x-3)/(x^2-2x+5)dx+int2/(x^2-2x+5)dx#

For:

#-int(3x-3)/(x^2-2x+5)dx#

Use the substitution:

#u=x^2-2x+5#

#implies du = 2x-2dx implies 3/2du = 3x-3dx#

#therefore -int(3x-3)/(x^2-2x+5)dx=-int(3/2)/udu=-3/2ln(u)+C#

Reverse the substitution:

#-3/2ln(x^2-2x+5)+C#

Now for the other integral:

#int2/(x^2-2x+5)dx#

Write the denominator in completed square form:

#x^2-2x+5=(x-1)^2-(-1)^2+5=(x-1)^2+4#

So:

#int2/(x^2-2x+5)dx=2intdx/((x-1)^2+4)#

Now substitute:

#2u = (x-1)#

#implies du=2dx# So:

#2intdx/((x-1)^2+4)=2int2/(4u^2+4)du=4/4int1/(u^2+1)du#

Which we recognize will simply integrate to inverse tangent giving us:

#=tan^-1(u)+C'#

Reverse the substitution:

#=tan^-1((x-1)/2)+C'#

Hence, the "something" is:

#int (-3x+5)/(x^2-2x+5)dx#

#=-int(3x-3)/(x^2-2x+5)dx+int2/(x^2-2x+5)dx#

#=-3/2ln(x^2-2x+5)+tan^-1((x-1)/2)+C#