Advertisements
Advertisements
प्रश्न
Evaluate the following integral:
`int ("d"x)/(2 - 3x - 2x^2)`
Advertisements
उत्तर
`int ("d"x)/(2 - 3x - 2x^2)`
Consider `2 - 3x - 2x^2 = 2[1 - 3/2 x - x^2]`
= `2[1 - (x^2 + 3/2)]`
= `2[1 - ((x + 3/4)^2 - 9/16)]`
= `2[1 + 9/16 - (x + 3/4)^2]`
= `2[25/16 - (x + 3/4)^2]`
= `2[(5/4)^2 - (x + 3/4)^2]`
So integral becomes,
`1/2 int ("d"x)/((5/4)^2 - (x + 3/4)^2) = 1/2 1/(2(5/4)) log |(5/4 + x + 3/4)/(5/4 - x - 3/4)|`
= `1/5 log|(2 + x)/(1 - 2x)| + "c"`
APPEARS IN
संबंधित प्रश्न
Integrate the following with respect to x.
`(sqrt(2x) - 1/sqrt(2x))^2`
Integrate the following with respect to x.
`x^3/(x + 2)`
Integrate the following with respect to x.
`(3x + 2)/((x - 2)(x - 3))`
Integrate the following with respect to x.
`("e"^x + 1)^2 "e"^x`
Integrate the following with respect to x.
`1/(x(log x)^2`
Integrate the following with respect to x.
sin3x
Integrate the following with respect to x.
`sqrt(1 - sin 2x)`
Choose the correct alternative:
`int ("d"x)/sqrt(x^2 - 36) + "c"`
Choose the correct alternative:
`int (2x + 3)/sqrt(x^2 + 3x + 2) "d"x` is
Choose the correct alternative:
`int_0^4 (sqrt(x) + 1/sqrt(x)) "d"x` is
