Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Advertisements
उत्तर
Let I =`int ("x"/("x + 1")^2) "e"^"x"` dx
`= int "e"^"x" ((("x + 1") - 1)/("x + 1")^2)` dx
`= int "e"^"x"(("x + 1")/("x + 1")^2 - 1/("x + 1")^2)` dx
`= int "e"^"x" (1/("x + 1") - 1/("x + 1")^2)` dx
Put f(x) = `1/("x + 1")`
∴ f '(x) = `(-1)/("x + 1")^2`
∴ I = `int "e"^"x" ["f"("x") + "f" '("x")]` dx
`= "e"^"x" * "f"("x") + "c"`
∴ I = `"e"^"x" (1/("x + 1"))` + c
Notes
The answer in the textbook is incorrect.
APPEARS IN
संबंधित प्रश्न
Integrate : sec3 x w. r. t. x.
Integrate the function in x log x.
Integrate the function in e2x sin x.
Evaluate the following : `int cos(root(3)(x)).dx`
Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
Choose the correct options from the given alternatives :
`int tan(sin^-1 x)*dx` =
Choose the correct options from the given alternatives :
`int sin (log x)*dx` =
Solve the following differential equation.
(x2 − yx2 ) dy + (y2 + xy2) dx = 0
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
`int 1/sqrt(2x^2 - 5) "d"x`
`int (cos2x)/(sin^2x cos^2x) "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
`int "e"^x x/(x + 1)^2 "d"x`
`int log x * [log ("e"x)]^-2` dx = ?
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
Evaluate the following:
`int_0^1 x log(1 + 2x) "d"x`
The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1) dx` is
Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`
Find: `int e^x.sin2xdx`
`int 1/sqrt(x^2 - a^2)dx` = ______.
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
`int1/sqrt(x^2 - a^2) dx` = ______
Evaluate the following.
`int x^3 e^(x^2) dx`
`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`
`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`
Evaluate:
`int1/(x^2 + 25)dx`
If ∫(cot x – cosec2 x)ex dx = ex f(x) + c then f(x) will be ______.
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
