Question

Evaluate the following integrals: 

\[\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}dx\]
\[\text{  We express  }x + 2 = A\left( \frac{d}{d x}\left( x^2 + 2x + 3 \right) \right) + B\]
\[x + 2 = \text{  A(2x + 2) }+ B\]
\[\text{Equating the coefficients of x and constants, we get}\]
\[\text{  1 = 2A and 2 = 2A + B }\]
\[\text{ or A }= \frac{1}{2} \text{ and B = 1}\]
\[ \therefore I = \int\frac{\frac{1}{2}\left( 2x + 2 \right) + 1}{\sqrt{x^2 + 2x + 3}}dx\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 2 \right)}{\sqrt{x^2 + 2x + 3}}dx + \int\frac{1}{\sqrt{x^2 + 2x + 3}}dx\]
\[ = \frac{1}{2} I_1 + I_2 . . . (1)\]
\[\text{ Now, }I_1 = \int\frac{\left( 2x + 2 \right)}{\sqrt{x^2 + 2x + 3}}dx\]
\[ \text{ Let }x^2 + 2x + 3 = u\]
` \text{ On differentiating both sides, we get `
\[ \left( 2x + 2 \right)dx = du\]
\[ \therefore I_1 = \int\frac{1}{\sqrt{u}}du\]
\[ = 2\sqrt{u} + c_1 \]
\[ = 2\sqrt{x^2 + 2x + 3} + c_1 . . . (2)\]
\[\text{ And,} I_2 = \int\frac{1}{\sqrt{x^2 + 2x + 3}}dx\]
\[ = \int\frac{1}{\sqrt{x^2 + 2x + 1 - 1 + 3}}dx\]
\[ = \int\frac{1}{\sqrt{\left( x + 1 \right)^2 + \left( \sqrt{2} \right)^2}}dx\]
\[ \text{ Let}\left( x + 1 \right) = u\]
`   \text{ On differentiating both sides, we get `
\[ dx = du\]
\[ \therefore I_2 = \int\frac{1}{\sqrt{\left( u \right)^2 + \left( \sqrt{2} \right)^2}}du\]
\[ = \text{ log}\left| u + \sqrt{\left( u \right)^2 + \left( \sqrt{2} \right)^2} \right| + c_2 \]
\[ = \text{ log}\left| \left( x + 1 \right) + \sqrt{x^2 + 2x + 3} \right| + c_2 . . . (3)\]
\[\text{ From (1), (2) and (3), we get}\]
\[ \therefore I = \frac{1}{2}\left( 2\sqrt{x^2 + 2x + 3} + c_1 \right) + \text{ log}\left| \left( x + 1 \right) + \sqrt{x^2 + 2x + 3} \right| + c_2 \]
\[ = \sqrt{x^2 + 2x + 3} + \text{ log  }\left| \left( x + 1 \right) + \sqrt{x^2 + 2x + 3} \right| + c\]
\[\text{ Hence, }\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}}dx = \sqrt{x^2 + 2x + 3} + \log\left| \left( x + 1 \right) + \sqrt{x^2 + 2x + 3} \right| + c\]