# Solving $\int \frac{8x+x^2}{(x-7)(x^2+16)} dx$

I am attempting to solve the following integral: $$\int \frac{8x+x^2}{(x-7)(x^2+16)} dx$$

Which is a bit hairy but great practice. However, my answer disagrees with online calculators so I need some guidance with it (it will be defined as Integral 1).

My answer is $2\arctan\ (\frac{x}{4})-47/32\ ln(x^2+16)+C$

It stems from equivocating the primary integral to: (defined as Integral 2)

$$\int \frac{(-47/16)\ x +8}{x^2+16} + \frac{63/16}{x-7} dx$$

I believe I integrated that properly, but to prove whether that integral and the first one written above are equivalent I have to show my method of finding its equivocation using partial fractions. I believe this is where my problem stems:

$$\frac{8x+x^2}{(x-7)(x^2+16)}=\frac{Ax+B}{x^2+16}+\frac{C}{x-7}$$

$$8x+x^2 = (Ax+B)\ (x-7) + C\ (x^2+16)$$ $$8x+x^2 = Ax^2 -7A + Bx -7B + Cx^2+16C$$

And then, since the RHS is a summation, the following statements must be true:

$$1 = A + C$$

$$8 = B$$

$$0 = -7A - 7B + 16C$$

This binds $8$ to $B$ and $C = 1-A$

Thus, for the third equation we have:

$$0 = -7A -56 + 16 - 16A$$

$$-23A = 40$$

$$A = -40/23$$

$$B=8$$

$$C= 63/23$$

It seems I've actually just gotten a different answer for the answers to my constants than in what I thought I had my constants as in Integral 2... Er, regardless, this does not agree with what WolframAlpha got:

$$\frac{8x+x^2}{(x-7)(x^2+16)}=\frac{-8(x-6)}{x^2+16}+\frac{21}{13(x-7)}$$

Can someone explain to me where I'm going wrong?

Maybe try this way Take $$\frac{x^2+8x}{x^3-7x+16x-112}$$ If you factor the denominator you get $(x-7)(x^2+16)$ Then for computing the partial sums: $$\frac{x^2+8x}{x^3-7x+16x-112} = \frac{A}{x-7}+\frac{Bx+Γ}{x^2+16}$$
You multiply by ${x^3-7x+16x-112}$ and you get $$x^2 +8x=A(x^2-16)+(Bx+Γ)(x-7)$$ After some operations: $$x^2+8x=Ax^2+Bx^2-7Bx+Γx+16A-7Γ$$ Then by comparing you get the equations: $$A+B=1$$ $$Γ-7B=8$$ $$16A-7Γ=0$$ Then by solving the matrix you get $$A=21/13$$ $$B=-8/13$$ $$Γ=48/13$$