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=?)?

2 Answers
Jan 28, 2018

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)

Jan 28, 2018

=-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