$\frac{ ? }{ {x}^{2}+7 }+\frac{ ? }{ x-2 }$
Since the factor in the denominator is quadratic, the numerator is an unknown linear expression $Ax+B$$\frac{ Ax+B }{ {x}^{2}+7 }+\frac{ ? }{ x-2 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $C$$\frac{ Ax+B }{ {x}^{2}+7 }+\frac{ C }{ x-2 }$
To get the unknown values, set the sum of fractions equal to the original fraction$\frac{ {x}^{2}-x-13 }{ \left( {x}^{2}+7 \right) \times \left( x-2 \right) }=\frac{ Ax+B }{ {x}^{2}+7 }+\frac{ C }{ x-2 }$
Multiply both sides of the equation by $\left( {x}^{2}+7 \right) \times \left( x-2 \right)$${x}^{2}-x-13=\left( x-2 \right) \times \left( Ax+B \right)+\left( {x}^{2}+7 \right)C$
Simplify the expression${x}^{2}-x-13=A{x}^{2}+Bx-2Ax-2B+C{x}^{2}+7C$
Use the commutative property to reorder the terms${x}^{2}-x-13=A{x}^{2}+C{x}^{2}+Bx-2Ax-2B+7C$
Group the powers of the $x$-terms and the constant terms${x}^{2}-x-13=\left( A+C \right){x}^{2}+\left( B-2A \right)x+\left( -2B+7C \right)$
When two polynomials are equal, their corresponding coefficients must be equal$\left\{\begin{array} { l } -13=-2B+7C \\ -1=B-2A \\ 1=A+C\end{array} \right.$
Solve the system of equations$\left( A, B, C\right)=\left( 2, 3, -1\right)$
Substitute the given values into the formed partial-fraction decomposition$\frac{ 2x+3 }{ {x}^{2}+7 }+\frac{ -1 }{ x-2 }$
Use $\frac{ -a }{ b }=\frac{ a }{ -b }=-\frac{ a }{ b }$ to rewrite the fraction$\frac{ 2x+3 }{ {x}^{2}+7 }-\frac{ 1 }{ x-2 }$