Integrals — Class 12 Maths Solution

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,$I = 3\sqrt {{x^2} + 9} + {C_1} - \log \left| {x + \sqrt {{x^2} + 9} } \right| - {C_2}$