Question
Evaluvate the following intregals:
$\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
$\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
Consider,
$5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+10)+\text{B}$
$\Rightarrow5\text{x}+3=\text{A}(2\text{x}+4)+\text{B}$
$\Rightarrow5\text{x}+3=(2\text{A})\text{x}+4\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=5$
$\Rightarrow\text{A}=\frac{5}{2}$
And
$4\text{A}+\text{B}=3$
$\Rightarrow4\times\frac{5}{2}+\text{B}=3$
$\Rightarrow\text{B}=-7$
$\therefore\text{I}=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+4-4+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{(\text{x}+2})^2+(\sqrt{6})^2}$
Putting, $\text{x}^2+4\text{x}+10=\text{t}$
$\Rightarrow(2\text{x}+4)\text{dx}=\text{dt}$
$\text{I}=\frac{5}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}-7\log\big|(\text{x}+2)^2+6\big|+\text{C}$
$=\frac{5}{2}\int\text{t}^{-\frac{1}{2}}\text{dt}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=\frac{5}{2}\times2\sqrt{\text{t}}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
Consider,
$5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+10)+\text{B}$
$\Rightarrow5\text{x}+3=\text{A}(2\text{x}+4)+\text{B}$
$\Rightarrow5\text{x}+3=(2\text{A})\text{x}+4\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=5$
$\Rightarrow\text{A}=\frac{5}{2}$
And
$4\text{A}+\text{B}=3$
$\Rightarrow4\times\frac{5}{2}+\text{B}=3$
$\Rightarrow\text{B}=-7$
$\therefore\text{I}=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+4-4+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{(\text{x}+2})^2+(\sqrt{6})^2}$
Putting, $\text{x}^2+4\text{x}+10=\text{t}$
$\Rightarrow(2\text{x}+4)\text{dx}=\text{dt}$
$\text{I}=\frac{5}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}-7\log\big|(\text{x}+2)^2+6\big|+\text{C}$
$=\frac{5}{2}\int\text{t}^{-\frac{1}{2}}\text{dt}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=\frac{5}{2}\times2\sqrt{\text{t}}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
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