Question

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

Answer 1

 

\[\text{ Let I }= \int\frac{x dx}{\sqrt{x^2 + 6x + 10}}\]
\[x = \text{ A  }\frac{d}{dx} \left( x^2 + 6x + 10 \right) + B\]
\[x = A \left( 2x + 6 \right) + B\]
\[x = \left( 2A \right) x + 6A + B\]
\[\text{Equating Coefficients of like terms}\]
\[2A = 1\]
\[A = \frac{1}{2}\]
\[6A + B = 0\]
\[6 \times \frac{1}{2} + B = 0\]
\[B = - 3\]
`  I  =  ∫  {x     dx}/{\sqrt{x^2 + 6x + 10}} `
\[ = \int\left( \frac{\frac{1}{2} \left( 2x + 6 \right) - 3}{\sqrt{x^2 + 6x + 10}} \right)dx\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 6 \right) dx}{\sqrt{x^2 + 6x + 10}} - 3\int\frac{dx}{\sqrt{x^2 + 6x + 3^2 - 3^2 + 10}}\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 6 \right) dx}{\sqrt{x^2 + 6x + 10}} - 3\int\frac{dx}{\sqrt{\left( x + 3 \right)^2 + 1^2}}\]
\[\text{ let x }^2 + 6x + 10 = t\]
\[ \Rightarrow \left( 2x + 6 \right) dx = dt\]
\[I = \frac{1}{2}\int\frac{dt}{\sqrt{t}} - 3\int\frac{dx}{\sqrt{\left( x + 3 \right)^2 + 1}}\]
\[ = \frac{1}{2} \times 2\sqrt{t} - 3 \text{ log }\left| x + 3 + \sqrt{\left( x + 3 \right)^2 + 1} \right| + C\]
\[ = \sqrt{t} - 3 \text{ log }\left| x + 3 + \sqrt{x^2 + 6x + 10} \right| + C\]
\[ = \sqrt{x^2 + 6x + 10} - 3 \text{ log } \left| x + 3 + \sqrt{x^2 + 6x + 10} \right| + C\]