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Let \[ I = \int {\frac{{3x – 1}}{{\sqrt {{x^2} + 9} }}} dx\]
\[I = \int {\frac{{3x}}{{\sqrt {{x^2} + 9} }}} dx – \int {\frac{1}{{\sqrt {{x^2} + 9} }}} dx\]
\[ I = {I_1} – {I_2}\]
Now, \[ {I_1} = \int {\frac{{3x}}{{\sqrt {{x^2} + 9} }}} \]
Let’s put \[ {x^2} + 9 = {t^2} \Rightarrow 2xdx = 2tdt \Rightarrow xdx = tdt\]
therefore,\[{I_1} = 3\int_t^t d t\]
\[ = 3\int d t = 3t + {C_1} = 3\sqrt {{x^2} + 9} + {C_1}\]
and \[ {I_2} = \int {\frac{1}{{\sqrt {{x^2} + 9} }}} dx = \int {\frac{1}{{\sqrt {{x^2} + {{(3)}^2}} }}} dx\]
\[ = \log \left| {x + \sqrt {{x^2} + 9} } \right| + {C_2}\]
therefore,\[ I = 3\sqrt {{x^2} + 9} + {C_1} – \log \left| {x + \sqrt {{x^2} + 9} } \right| – {C_2}\]
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