Question

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

`int {3x^2+5x+50}/{x^4+x^2+10x}dx=...=int (5/x-2/{x+2}+{-3x+5}/{x^2-2x+5})dx=5ln|x|-2ln|x+2|+SOMETHING+c`
whtat is the "something" (`int {-3x+5}/{x^2-2x+5}dx=?`)?

Answer 1

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

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

Explanation:

`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)`

Answer 2

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

Explanation:

`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`