Question

फलन का समाकलन कीजिए।

`1/sqrt(1 + 4x^2)`

Answer 1

माना `I = int 1/sqrt(1 + 4x^2)  dx`

`= 1/2 int dx/sqrt(1/4 + x)`

`= 1/2 int dx/sqrt((1/2)^2 + x^2)`

`= 1/2 log abs (x sqrt(1/4 +x^2)) C_1`

       .....`[because int dx/sqrt(x^2 + a^2) = log abs (x + sqrt(x^2 + a^2)) + C]`

`= 1/2 log abs ((2x + sqrt(1 + 4x^2))/2) + C_1`

`= 1/2 log abs (2x + sqrt(1 + 4x^2)) - 1/2 log 2 + C_1`

`= 1/2 log |2x + sqrt 1 + 4x^2| + C`      .... [`C = -1/2 log 2 + C_1`]

Answer 2

माना `I = int 1/(sqrt(1 + 4x^2)) dx = int dx/sqrt(1 + (2x)^2)`

`2x + tan theta` रखने पर,

`2  dx = sec^2 theta  d theta`

`dx = 1/2 sec^2 theta  d theta`

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

`= 1/2 int (sec^2 theta)/(sec theta)  d theta`

`= 1/2 int sec theta  d theta`

`= 1/2  log (sec theta + tan theta) + C`

`= 1/2  log [tan theta + sqrt(1 + tan^2 theta)] + C`

`= 1/2  log (2x + sqrt(1 + (2x)^2) + C)`

`= 1/2  log [2x + sqrt(1 + 4x^2)] + C`