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.
(3 + x)(2 – 5x)
Integrate the following with respect to x.
`1/(sqrt(x + 1) + sqrt(x - 1))`
Integrate the following with respect to x.
`(x^3 + 3x^2 - 7x + 11)/(x + 5)`
Integrate the following with respect to x.
`("e"^(3x) - "e"^(-3x))/"e"^x`
Integrate the following with respect to x.
`("e"^(3x) +"e"^(5x))/("e"^x + "e"^-x)`
Integrate the following with respect to x.
2 cos x – 3 sin x + 4 sec2x – 5 cosec2x
Integrate the following with respect to x.
x log x
Integrate the following with respect to x.
xn log x
Integrate the following with respect to x.
`("e"^(3logx))/(x^4 + 1)`
Choose the correct alternative:
`int logx/x "d"x, x > 0` is
