Mathematics | TS EAMCET 2018 | Discussion Page

$\frac{d}{dx}\left(\frac{x+5}{(x+1)^{2}(x+2)}\right)=$

$\frac{8}{(x+2)^{2}}-\frac{3}{(x+1)^{2}}+\frac{3}{(x+1)^{3}}$

$\frac{3}{(x+1)^{2}}-\frac{3}{(x+2)^{2}}-\frac{8}{(x+1)^{3}}$

$\frac{3}{(x+2)^{2}}-\frac{3}{(x+1)^{3}}-\frac{8}{(x+1)^{2}}$

$\frac{8}{(x+2)^{2}}-\frac{3}{(x+1)^{3}}+\frac{3}{(x+1)^{2}}$

Option B

$\frac{x+5}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)^{}}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)^{}}$ ..........(i)

$\Rightarrow$ $x+5= A(x+1)(x+2)+B(x+2)+c(x+1)^{2}$

$\Rightarrow$ $x+5= x^{2}(A+C)+x(3A+B+2C)+2A+2B+C$

on comparing the coefficients of like power of x , we get

$A+C=0$ ,

$3A+B+2C=1$ and $2A+2B+C=5$

on substituting A=-C in last two equations, we get

B-C=1 .....(ii)

and 2B-C=5 .......(iii)

On subtracting Eqs(ii) and (iii) , we get

$B=4 \Rightarrow C=3$ [from Eq.(ii)]

and hence A=-3

Now, $\frac{x+5}{(x+1)^{2}(x+2)}=\frac{-3}{(x+1)^{}}+\frac{4}{(x+1)^{2}}+\frac{3}{(x+2)^{}}$

On differentiating both sides w.r.t x, we get

$\frac{d}{dx}\left(\frac{x+5}{(x+1)^{2}(x+2)}\right)=$ $\frac{3}{(x+1)^{2}}-\frac{8}{(x+1)^{3}}-\frac{3}{(x+2)^{2}}$

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