x = a(x<sup>2</sup> +9) + (bx +c)(x+1)
= (a+b)x<sup>2</sup> + (b+c)x + 9a + c
a+b=0
b+c=1
9a+c=0
solve simultaneously and a= -1/10, b= 1/10, c=9/10 so you get:
1/10 ∫ x/(x<sup>2</sup> +9) + 9/(x<sup>2</sup> +9) - 1/(x+1) dx
= 1/10[(1/2)ln(x<sup>2</sup> +9) + 3tan<sup>-1</sup>(x/3) -ln(x+1)] + C