Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int "x"^2 "e"^"4x"`dx
उत्तर
Let I = `int "x"^2 "e"^"4x"`dx
`= "x"^2 int "e"^"4x" "dx" - int["d"/"dx" ("x"^2) int "e"^"4x" "dx"]` dx
`= "x"^2 * "e"^"4x"/4 - int 2"x" * "e"^"4x"/4` dx
`= ("x"^2 * "e"^"4x")/4 - 1/2 int "x" * "e"^"4x"` dx
`= ("x"^2 * "e"^"4x")/4 - 1/2 ["x" int "e"^"4x" "dx" - int ("d"/"dx" ("x") int "e"^"4x" "dx") "dx"]`
`= ("x"^2 * "e"^"4x")/4 - 1/2 ["x" * "e"^"4x"/4 - int 1 * "e"^"4x"/4 "dx"]`
`= ("x"^2 "e"^"4x")/4 - 1/2[("x" * "e"^"4x")/4 - 1/4 int "e"^"4x" "dx"]`
`= ("x"^2 "e"^"4x")/4 - 1/2[("x" * "e"^"4x")/4 - 1/4 * "e"^"4x"/4]` + c
`= ("x"^2 "e"^"4x")/4 - ("x" "e"^"4x")/8 + "e"^"4x"/32` + c
∴ I = `("e"^"4x")/4 ["x"^2 - "x"/2 + 1/8]` + c
Notes
The answer in the textbook is incorrect.
APPEARS IN
संबंधित प्रश्न
Prove that:
`int sqrt(a^2 - x^2) dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a)+c`
Integrate the function in (x2 + 1) log x.
Evaluate the following : `int cos sqrt(x).dx`
Integrate the following functions w.r.t. x:
sin (log x)
Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`
Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Choose the correct options from the given alternatives :
`int (1)/(cosx - cos^2x)*dx` =
Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
Choose the correct alternative from the following.
`int (1 - "x")^(-2) "dx"` =
Choose the correct alternative from the following.
`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5 "dx"` =
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
`int sqrt(tanx) + sqrt(cotx) "d"x`
`int log x * [log ("e"x)]^-2` dx = ?
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
If `int (f(x))/(log(sin x))dx` = log[log sin x] + c, then f(x) is equal to ______.
Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))dx`.
Find `int e^x ((1 - sinx)/(1 - cosx))dx`.
Evaluate `int(3x-2)/((x+1)^2(x+3)) dx`
Evaluate:
`intcos^-1(sqrt(x))dx`
