Need help splitting (x + 1) / (6x + x^2) into partial fractions?

(x + 1) / (6x + x²) = (x + 1) / x (x + 6)

Let (x + 1) / (6x + x²) = (A / x) + [B / (x + 6)]

Hence, (A / x) + [B / (x + 6)] = (x + 1) / [x (x + 6)]
[A(x + 6) / x(x + 6)] + [B x / x(x + 6)] = (x + 1) / [x (x + 6)]
[A(x + 6) + Bx] / x(x + 6) = (x + 1) / [x (x + 6)]

As the denominators are equal, then the numerators should be equal.
A(x + 6) + Bx = x + 1

Put x = 0 :
6A = 1
A = 1/6

Put x = -6 :
-6B = -5
B = 5/6

Hence, (x + 1) / (6x + x²) = (1 / 6x) + [5 / 6(x + 6)]

(x + 1) / [ x (6 + x) ] = A / x + B / [ 6 + x ]
x + 1 = A (6 + x) + B x
1 = A + B
1 = 6A

A = 1/6
B = 5/6

(x + 1) / [ x (6 + x) ] = 1 / [6x] + 5 / [ 36 + 6x ]

// factor denominator
(x+1)/(x²+6x) = (x+1)/[x(x+6)]
= A/x + B/(x+6)

// common denominator
// numerator is
A(x+6) + B(x) = x+1

// choose convenient values of x
let x=0, 6A = 1, A = 1/6
let x=-6, -6B = -5, B = 5/6

== 1/6 [1/x + 5/(x+6)]

check

1/6 (x+6+5x)/[x(x+6)] =
(1/6)(6x+6) / [x(x+6)] = (x+1)/[x(x+6)]

Well,

(x+1)/(6x + x^2) = (x+1)/ [ x(x+6) ]
therefore, there exist a and b so that
(x+1)/(6x + x^2) = a/x + b/(x+6)
= [ a(x + 6) + bx ] /(6x + x^2)
= [ ax + 6a + bx ] /(6x + x^2)
= [ (a+b)x + 6a ] /(6x + x^2)
therefore, we must have :
a + b = 1
and
6a = 1 ==> a = 1/6
so
a=1/6 and a+b = 1 ==> b = 5/6
conclusion :

(x+1)/(6x + x^2) = 1/6 * (1/x + 5/(x+6) )

hope it' ll help !!