Integration?

Question

integrate:
1. 10/ ((x-1)(x^2+9)) dx
2. (x+4)/(x^2+2x+5) dx

Answer 1

1) This is a partial fractions question.
Let 10/ ((x-1)(x^2+9)) = (Ax+B)/(x^2+9) + C/(x-1)
So (Ax+B)(x-1) + C(x^2+9) = 10
When x=1 then 10C=10 so C=1
If we compare coefficients of x^2 then Ax^2+Cx^2=0 so A = -1
When x=0 then -B + 9C = 10 so B = -1

So 10/ ((x-1)(x^2+9))
= -(x+1)/(x^2+9) +1/(x-1)

So ∫10/ ((x-1)(x^2+9)) dx
=∫-(x+1)/(x^2+9) +1/(x-1) dx
=-∫x/(x^2+9) dx - ∫1/(x^2+9) dx +∫1/(x-1) dx
= -1/2 ln(x^2+9) - 1/3arctan(x/3) + ln|x-1| + c

2) First lets complete the square on the denominator:
(x^2+2x+5) = (x+1)^2+4
Lets now rewrite the integral as:
∫(x+1+3)/((x+1)^2+4) dx
=∫(x+1)/((x+1)^2+4) dx+∫(3)/((x+1)^2+4) dx
=1/2 ln((x+1)^2+4) + 3/2arctan((x+1)/2) +c

Answer 2

1) [6ln(|x-1|)-3ln(x^2+9)-arctan(x/3)]/6+c
2) [ln(|x^2+2x+5|)+3arctan((x+1)/2)]/2+c