∫x2+a2dxLet x=atanudx=asec2udu=∫a2tan2u+a2asec2udu=a∫a2(1+tan2u)sec2udu⇒aa∫secusec2udu=∫secusec2udu=∫secudu=ln[tan(u)+sec(u)]+c[∵∫secxdx=tanx]=ln[a2x+1+a2x2]+c=ln[ax+aa2+x2]+c=ln[ax+x2+a2]+c\therefore \quad$ Option (A) is correct.