`int sqrt((2x^(2)-5x-1)) dx` का समाकलन कीजिए।

Answer

.

Explanation

`int sqrt(2x^(2)-5x-1) dx`
`=int sqrt(2) sqrt(x^(2)-5/2x-1/2) dx = sqrt(2) int sqrt(x^(2)-5/2 x+25/16-25/16-1/2) dx`
`-sqrt(2) int sqrt((x-5/4)^(2)-33/16) dx=sqrt(2) int sqrt((x-5/4)^(2)-(sqrt(33)/4)^(2)) dx`,
सूत्र `int sqrt(x^(2)-a^(2)) dx=1/2 xsqrt(x^(2)-a^(2))-1/2 a^(2) log (x+sqrt(x^(2)-a^(2))` से
`=sqrt(2) [1/2 (x-5/4) sqrt((x-5/4)^(2)-(sqrt(33)/4)^(2))-1/2 33/16 log {(x-5/4)+sqrt((x-5/4)^(2)-(sqrt(33)/4)^(2))}]+c`
`=sqrt(2) 1/2 [(x-5/4) sqrt(x^(2)+25/16-5/2x-33/16)-33/16log {(x-5/4)+sqrt(x^(2)+25/16-5/2x-33/16)}]+c`
`=1/sqrt(2) [(x-5/4)sqrt((16x^(2)-40x-8)/16)-33/16 log {(x-5/4)+sqrt((16x^(2)-40x-8)/16)}]+c`
`=1/sqrt(2) [(x-5/4) sqrt((2x^(2)-5x-1)/2)-33/16 log {(x-5/4)+sqrt((2x^(2)-5x-1)/2)}]+c`
`=1/sqrt(2) [(x-5/4) 1/sqrt(2) sqrt(2x^(2)-5x-1) -33/16 log {(x-5/4)+1/sqrt(2) sqrt(2x^(2)-5x-1)}]+c`
`=1/2 (x-5/4) sqrt(2x^(2)-5x-1) -33/(16sqrt(2)) log {(x-5/4)+1/sqrt(2) sqrt(2x^(2)-5x-1)}+c`.