Answer:
[tex]\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{3}{x+1}+\dfrac{4}{x+5}[/tex]
Explanation:
The given expression is
[tex]\dfrac{7x+19}{(x+1)(x+5)}[/tex]
We need to resolve this into partial fraction.
The form of the partial fraction decomposition is
[tex]\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{A}{x+1}+\dfrac{B}{x+5}[/tex] ...(1)
[tex]\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{A(x+5)+B(x+1)}{(x+1)(x+5)}[/tex]
[tex]7x+19=Ax+5A+Bx+B[/tex]
[tex]7x+19=(A+B)x+(5A+B)[/tex]
On comparing both sides, we get
[tex]A+B=7[/tex] ...(2)
[tex]5A+B=19[/tex] ...(3)
Subtract (2) from (3), we get
[tex]4A=12[/tex]
[tex]A=3[/tex]
Put A=3 in (1).
[tex]3+B=7[/tex]
[tex]B=4[/tex]
Put A=3 and B=4 in (1).
[tex]\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{3}{x+1}+\dfrac{4}{x+5}[/tex]
Therefore, [tex]\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{3}{x+1}+\dfrac{4}{x+5}[/tex] .