Question

what's the solution??​

Answer 1

       $\underline{\bold{Answer:}}$

       $\boxed{2(\sqrt{x^{2} + 4x + 3}) +log[\frac{1}{x^{2} + 5x + 5} ] +C}$

       $\underline{\bold{Step-by-step-explaination:}}$

       $\text{The question states:}$

       $\boxed{\int{\frac{2x+3}{\sqrt{(x+1)(x+3)}} } \, dx = I}$

       $\text{We can write }(2x + 3 )\text{ as }(2x+4-1)$

$\implies \boxed{\int{\frac{2x+4-1}{\sqrt{(x+1)(x+3)}} } \, dx = I}$

       $\text{We know that }\int{(ax - bx})dx = \int{ax}dx-\int{bx}dx$

$\implies \boxed{\int{\frac{2x-4}{\sqrt{(x+1)(x+3)}} }dx - \int{\frac{1}{\sqrt{(x+1)(x+3)}} }dx=I}$

       $\text{Let } \boxed{\int{\frac{2x+4}{\sqrt{(x+1)(x+3)}} } \, dx = I_1} \text{ and } \boxed{-\int{\frac{1}{\sqrt{(x+1)(x+3)}} } \, dx = I_2}$

$\implies \boxed{I = I_1 + I_2}$

$\implies \boxed{I_1=\int{\frac{2x+4}{\sqrt{x^{2} + 4x + 3 }} } \, dx }$

       $\text{Let }(x^{2} + 4x + 3) = t^{2}$

       $\text{Differentiating both sides}$

$\implies \boxed{2tdt = (2x + 4)dx}$

$\implies \boxed{I_1=\int{\frac{2t}{\sqrt{t^{2}}}dt}}$

$\implies \boxed{I_1=\int{\frac{2t}{t}}dt}}$

$\implies \boxed{I_1=\int{2}dt}}$

       $\text{Applying the constant rule}$

$\implies \boxed{I_1=2\int{1}dt}}$

$\implies \boxed{I_1=2t + C_1}$

       $\text{We know that }\boxed{t^{2} = x^{2} + 4x + 3}$

$\implies \boxed{t = \sqrt{x^{2} + 4x + 3}}$

$\implies \boxed{I_1 = 2(\sqrt{x^{2} + 4x + 3}) + C_1} \text{ ------(i)}$

$\implies \boxed{I_2 = {-\int{\frac{1}{\sqrt{(x+1)(x+3)}}} dx}}$

$\implies \boxed{I_2 = {-\int{\frac{1}{\sqrt{x^{2} + 4x + 3}}} dx}}$

       $\text{We can write }(x^{2} + 4x + 3) \text{ as }(x^{2} + 4x + 4 - 1)$

$\implies \boxed{I_2 = {-\int{\frac{1}{\sqrt{x^{2} + 4x + 4-1}}} dx}}$

$\implies \boxed{I_2 = {-\int{\frac{1}{\sqrt{(x+2)^{2} - 1}}} dx}}$

$\implies \boxed{I_2 = {-\int{\frac{1}{\sqrt{(x+2)^{2} - (1)^{2} }}} dx}}$

       $\text{Let }(x+2) = m$

$\implies \boxed{dm = 1dx}$

$\implies \boxed{I_2 = {-\int{\frac{dm}{\sqrt{(m)^{2} - (1)^{2} }}}}}$

       $\text{Applying }\int\frac{1}{\sqrt{y^{2} - a^{2}}} dy = log[y + (y^{2} - a^{2} )]$

$\implies \boxed{I_2= -log[m + (m^{2} - 1)]+ C_2}$

       $\text{Putting }m = (x+2) \text{ back into the equation}$

$\implies \boxed{I_2 = - log[x+2+(x^{2} + 4x + 3)]+C_2}$

$\implies \boxed{I_2 = - log[x^{2} + 5x + 5]+C_2}$

       $\text{We know that }a[log(b )]= log(b^{a} )$

$\implies \boxed{I_2 = log[(x^{2} + 5x + 5)^{-1}] +C_2}$

       $\text{We know that }a^{-1 }= \frac{1}{a}$

$\implies \boxed{I_2 = log[\frac{1}{x^{2} + 5x + 5} ] +C_2}\text{ ------(ii)}$

$\implies \boxed{I= I_1+I_2}$

       $\text{From (i) and (ii)}$

$\implies \boxed{I = 2(\sqrt{x^{2} + 4x + 3}) + C_1+log[\frac{1}{x^{2} + 5x + 5} ] +C_2}$

       $C_1 + C_2 = C$

$\implies \boxed{I = 2(\sqrt{x^{2} + 4x + 3}) +log[\frac{1}{x^{2} + 5x + 5} ] +C}$