Advertisements
Advertisements
प्रश्न
Evaluate: `int "e"^sqrt"x"` dx
उत्तर
Let I = `int "e"^sqrt"x"` dx
Put `sqrt"x"` = t
∴ x = t2
∴ dx = 2t dt
∴ I = `int "e"^"t" * "2t"`dt
`= 2 int "t" * "e"^"t" * "dt"`
`= 2 ["t" int "e"^"t" "dt" - int {"d"/"dx" ("t") int "e"^"t" * "dt"}"dt"]`
`= 2 ["t" * "e"^"t" - int 1 * "e"^"t" "dt"]`
`= 2("te"^"t" - "e"^"t")` + c
`= 2"e"^"t" ("t - 1")` + c
∴ I = `2"e"^sqrt"x" (sqrt"x" - 1)` + c
APPEARS IN
संबंधित प्रश्न
Integrate the functions:
`e^(tan^(-1)x)/(1+x^2)`
Write a value of\[\int\frac{1}{1 + 2 e^x} \text{ dx }\].
Evaluate the following integrals : `int (3)/(sqrt(7x - 2) - sqrt(7x - 5)).dx`
Integrate the following functions w.r.t. x : `((x - 1)^2)/(x^2 + 1)^2`
Integrate the following functions w.r.t. x : `(3e^(2x) + 5)/(4e^(2x) - 5)`
Evaluate the following:
`int (1)/(25 - 9x^2)*dx`
If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).
Evaluate the following.
`int "x" sqrt(1 + "x"^2)` dx
`int sqrt(1 + "x"^2) "dx"` =
`int ("x + 2")/(2"x"^2 + 6"x" + 5)"dx" = "p" int (4"x" + 6)/(2"x"^2 + 6"x" + 5) "dx" + 1/2 int "dx"/(2"x"^2 + 6"x" + 5)`, then p = ?
Evaluate:
`int (5x^2 - 6x + 3)/(2x − 3)` dx
Evaluate `int 1/((2"x" + 3))` dx
`int sqrt(x) sec(x)^(3/2) tan(x)^(3/2)"d"x`
`int (f^'(x))/(f(x))dx` = ______ + c.
Write `int cotx dx`.
Evaluated the following
`int x^3/ sqrt (1 + x^4 )dx`
If f ′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)
Evaluate.
`int (5x^2-6x+3)/(2x-3)dx`
Evaluate:
`int(cos 2x)/sinx dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
