Integrals — Maths STD 12 Science — Question

$= \int {\sqrt {\frac{{9 + {x^2}}}{9}} } dx$

$ = \int {\frac{{\sqrt {{x^2} + {3^2}} }}{3}} dx$

$= \frac{1}{3}\int {\sqrt {{x^2} + {3^2}} } dx$

$= \frac{1}{3}\left[ {\frac{x}{2}\sqrt {{x^2} + {3^2}} + \frac{{{3^2}}}{2}\log \left| {x + \sqrt {{x^2} + {3^2}} } \right|} \right] + c$

$\left[ {\because \int {\sqrt {{x^2} + {a^2}} dx} } \right.$ $\left. { = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + c} \right]$

$= \frac{x}{6}\sqrt {{x^2} + 9} + \frac{3}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right| + c$