Evaluate: $\int \frac{5 \mathrm{x} + 3}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \mathrm{dx}$

$$\begin{gathered} \int \frac{5 \mathrm{x} + 3}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \mathrm{dx} \\ \mathrm{Now}, 5 \mathrm{x} + 3 = \mathrm{A} \frac{\mathrm{d}}{\mathrm{dx}} \left({\mathrm{x}}^2 + 4 \mathrm{x} + 10 \right) + \mathrm{B} \\ \Rightarrow 5 \mathrm{x} + 3 = \mathrm{A} \left( 2 \mathrm{x} + 4 \right) + \mathrm{B} \\ \Rightarrow 5 \mathrm{x} + 3 = 2 \mathrm{A} \mathrm{x} + 4 \mathrm{A} + \mathrm{B} \\ \Rightarrow 2 \mathrm{A} = 5 \mathrm{and} 4 \mathrm{A} + \mathrm{B} = 3 \\ \Rightarrow \mathrm{A} = \frac{5}{2} \\ \mathrm{Thus}, 4 \left( \frac{5}{2} \right) + \mathrm{B} = 3 \\ \Rightarrow 10 + \mathrm{B} = 3 \\ \Rightarrow \mathrm{B} = 3 - 10 = -7 \end{gathered}$$

On substituting the values of  A  and  B,  we get

$$\begin{gathered} \int \frac{\left( 5 \mathrm{x} + 3 \right)}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \mathrm{dx} = \int \frac{\left[{\displaystyle \frac{5}{2}} {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}} \left({\mathrm{x}}^2 + 4 \mathrm{x} + 10 \right) - 7 \right]}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \mathrm{dx} \\ =\int \left[\frac{{\displaystyle \frac{5}{2} \left( 2 \mathrm{x} +4 \right) - 7 }}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \right] \mathrm{dx} \\ = \frac{5}{2} \int \frac{2 \mathrm{x} +4}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \mathrm{dx} - 7\int \frac{\mathrm{dx}}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \\ = \frac{5}{2} {\mathrm{I}}_1 - 7 {\mathrm{I}}_2 ...........( \mathrm{i} ) \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_2 = \int \frac{2 \mathrm{x} + 4}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \mathrm{dx} \\ \mathrm{Put} {\mathrm{x}}^2 + 4 \mathrm{x} + 10 = {\mathrm{z}}^2 \\ \left(2 \mathrm{x} + 4 \right) \mathrm{dx} = 2 \mathrm{z} \mathrm{dz} \\ \mathrm{Thus}, {\mathrm{I}}_1 = \int \frac{2 \mathrm{z}}{\mathrm{z}} \mathrm{dz} = 2 \mathrm{z} = 2 \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} + {\mathrm{C}}_1 \\ {\mathrm{I}}_2 = \int \frac{\mathrm{dx}}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} \\ = \int \frac{\mathrm{dx}}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 4 + 6}} \\ = \int \frac{\mathrm{dx}}{\sqrt{ ( \mathrm{x} + 2 {)}^2 + {\left( \sqrt{ 6} \right)}^2}} \\ = \log \left| ( \mathrm{x} + 2 ) + \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} \right| + {\mathrm{C}}_2 \end{gathered}$$

Substituting  I₁  and  I₂  in  ( i ),  we get

$$\begin{gathered} \therefore \int \frac{5 \mathrm{x} + 3}{\sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10}} = \frac{5}{2} \left( 2 \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} + {\mathrm{C}}_1 \right) - 7 \left[ \log \left| \left( \mathrm{x} + 2 \right) + \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} \right| + {\mathrm{C}}_2 \right] \\ = 5 \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} - 7 \left[ \log \left| \left( \mathrm{x} + 2 \right) + \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} \right| \right] + \frac{5}{2} {\mathrm{C}}_1 - 7 {\mathrm{C}}_2 \\ = 5 \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} - 7 \left[ \log \left| \left( \mathrm{x} + 2 \right) + \sqrt{ {\mathrm{x}}^2 + 4 \mathrm{x} + 10} \right| \right] +\mathrm{C}, \mathrm{where} \mathrm{C} = \frac{5}{2} {\mathrm{C}}_1 - 7 {\mathrm{C}}_2 \end{gathered}$$