Advertisements
Advertisements
Question
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
Advertisements
Solution
Let I = `int "e"^"x"/(4"e"^"2x" -1)` dx
`"I" = int "e"^"x"/(4("e"^"x")^2 - 1)` dx
Put ex = t
∴ ex dx = dt
∴ I = `int "dt"/(4"t"^2 - 1)`
`∴ "I" = 1/4 int 1/("t"^2 - 1/4)` dt
`∴ "I" = 1/4 int 1/("t"^2 - (1/2)^2)` dt
`∴ "I" = 1/4 . 1/(2 (1/2)) log |("t" - 1/2)/("t" + 1/2)|` + c
`∴ "I" = 1/4 log |("2t" - 1)/("2t" + 1)|` + c
Resubstitute t = ex
`∴ "I" = 1/4 log |(2"e"^"x" - 1)/(2"e"^"x" + 1)|` + c
Notes
Answer in the textbook is incorrect.
APPEARS IN
RELATED QUESTIONS
Integrate the function in (sin-1x)2.
Integrate the function in x sec2 x.
`int e^x sec x (1 + tan x) dx` equals:
Evaluate the following:
`int x tan^-1 x . dx`
Evaluate the following : `int x^3.logx.dx`
Integrate the following functions w.r.t.x:
`e^-x cos2x`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
Choose the correct options from the given alternatives :
`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =
Choose the correct options from the given alternatives :
`int (x- sinx)/(1 - cosx)*dx` =
If f(x) = `sin^-1x/sqrt(1 - x^2), "g"(x) = e^(sin^-1x)`, then `int f(x)*"g"(x)*dx` = ______.
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`
`int (sinx)/(1 + sin x) "d"x`
`int ("e"^xlog(sin"e"^x))/(tan"e"^x) "d"x`
`int ("d"x)/(x - x^2)` = ______
Choose the correct alternative:
`int ("d"x)/((x - 8)(x + 7))` =
`int(x + 1/x)^3 dx` = ______.
`int"e"^(4x - 3) "d"x` = ______ + c
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?
`int cot "x".log [log (sin "x")] "dx"` = ____________.
`int(1-x)^-2 dx` = ______
Evaluate `int(1 + x + (x^2)/(2!))dx`
The value of `int e^x((1 + sinx)/(1 + cosx))dx` is ______.
Evaluate the following:
`intx^3e^(x^2)dx`
The value of `inta^x.e^x dx` equals
Evaluate `int(1 + x + x^2/(2!))dx`.
