What is the integration of \[ \int {\frac{{3x – 1}}{{\sqrt {{x^2} + 9} }}} dx\] ~~~~~[NCERT Exemp. Q. 16,Page 164 ]

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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