Prove that: `int "dx"/(sqrt("x"^2 +"a"^2)) = log |"x" +sqrt("x"^2 +"a"^2) | + "c"`

#### Solution

Let I = `int 1/sqrt("x"^2 + "a"^2) "dx"`

Put x = a tan θ ⇒ tan θ = `"x"/"a"`

∴ dx = a sec^{2} θ dθ

∴ I = `int 1/ sqrt("a"^2 "tan"^2 theta +"a"^2) "a" "sec"^2 theta "d" theta`

= `int ("a"."sec"^2 theta)/("a" sqrt(1+"tan"^2 theta)) "d"theta`

= `int ("sec"^2 theta)/("sec" theta) "d"theta `

`= int "sec" theta . "d" theta`

`= "log" |"sec" theta +"tan" theta| +"c"_1`

`= "log" |"x"/"a" + sqrt("sec"^2 theta)| + "c"_1`

`= "log" | "x"/"a" + sqrt 1+ "tan"^2 theta | + "c"_1`

=`"log" |"x" /"a" +sqrt(1+"x"^2/"a"^2)| +"c"_1`

=` "log" |"x"/"a" + sqrt( "a"^2 + "x"^2)/"a"| + "c"_1`

`= "log" |"x" +sqrt("x"^2 +"a"^2)| - "log" "a" + "c"_1`

`therefore int 1/sqrt("x"^2 + "a"^2) "dx" = "log" |"x" +sqrt("x"^2 +"a"^2)| - "log" "a" + "c" ,`

where c = - log a +c_{1}