1.

2x+3/(x+3)(x^2+4)

Answer»

\(\frac{2x+3}{(x+3)(x^2+4)}=\frac{A}{x+3}+\frac{Bx+C}{x^2+4}\)

⇒ 2x + 3 = A(x2+4) + (Bx+C) (x+3)

⇒ (A+B)x2 + (3B+C)x + 4A+3C = 2x+3

By comparing corresponding components, we get

A+B = 0  ...(1)

3B+C = 2  ...(2)

4A+3C = 3  ...(3)

Put B = -A from equation (1) into equation (2), we get

-3A+C = 2

⇒ -9A+3C = 6  ...(4)

Subtract equation (4) from equation (3), we get

13A = -3

⇒ A = -3/13

∴ B = -A = 3/13

and C = 2 - 3B = 2 - 9/13 = 17/13

∴ \(\frac{2x+3}{(x+3)(x^2+4)}=\frac{-3/13}{x+3}+\cfrac{\frac{3}{13}x+\frac{17}{13}}{x^2+4}\)



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