Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int 1/(4"x"^2 - 1)` dx
Advertisements
उत्तर
Let I = `int "dx"/(4"x"^2 - 1)`
`= 1/4 int "dx"/("x"^2 - 1/4)`
`= 1/4 int "dx"/("x"^2 - (1/2)^2)`
`= 1/4 xx 1/(2 xx 1/2) log |("x" - 1/2)/("x" + 1/2)|` + c
∴ I = `1/4` log `|("2x" - 1)/("2x" + 1)|` + c
Alternate Method:
Let I = `int "dx"/(4"x"^2 - 1) = int "dx"/((2"x"^2) - (1)^2)`
`= 1/(2 xx 1) xx 1/2 log |("2x" - 1)/("2x" + 1)|` + c
∴ I = `1/4` log `|("2x" - 1)/("2x" + 1)|` + c
APPEARS IN
संबंधित प्रश्न
Evaluate :`intxlogxdx`
Find `int((3sintheta-2)costheta)/(5-cos^2theta-4sin theta)d theta`.
Integrate the functions:
tan2(2x – 3)
Evaluate `int (x-1)/(sqrt(x^2 - x)) dx`
Evaluate: `int (sec x)/(1 + cosec x) dx`
Integrate the following functions w.r.t. x : `(e^(2x) + 1)/(e^(2x) - 1)`
Integrate the following functions w.r.t. x:
`x^5sqrt(a^2 + x^2)`
Integrate the following functions w.r.t. x : `(1)/(x.logx.log(logx)`.
Integrate the following functions w.r.t. x : tan 3x tan 2x tan x
Evaluate the following integrals : `int sqrt((x - 7)/(x - 9)).dx`
Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`
To find the value of `int ((1 + log x) )/x dx` the proper substitution is ______.
Evaluate: `int 1/(2"x" + 3"x" log"x")` dx
Evaluate: `int (2"e"^"x" - 3)/(4"e"^"x" + 1)` dx
`int "dx"/((sin x + cos x)(2 cos x + sin x))` = ?
`int[ tan (log x) + sec^2 (log x)] dx= ` ______
`int (sin (5x)/2)/(sin x/2)dx` is equal to ______. (where C is a constant of integration).
Evaluate `int_(logsqrt(2))^(logsqrt(3)) 1/((e^x + e^-x)(e^x - e^-x)) dx`.
Solve the following Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3)dx`
Evaluate the following.
`int x sqrt(1 + x^2) dx`
Evaluate:
`intsqrt(3 + 4x - 4x^2) dx`
Evaluate `int(1+x+(x^2)/(2!))dx`
Evaluate `int(1+x+x^2/(2!))dx`
Evaluate:
`int(5x^2-6x+3)/(2x-3)dx`
Evaluate `int1/(x(x-1))dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
