हिंदी

Evaluate ∫ex(1x-1x2) dx

Advertisements
Advertisements

प्रश्न

Evaluate  `int"e"^x (1/x - 1/x^2)  "d"x`

योग
Advertisements

उत्तर

Let I = `int"e"^x (1/x - 1/x^2)  "d"x`

Put f(x) = `1/x`

∴ f'(x) = `-1/x^2`

∴ I = `int"e"^x ["f"(x) + "f'"(x)]  "d"x`

= `"e"^x*"f"(x) + "c"`

∴ I = `"e"^x* 1/x + "c"`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1.5: Integration - Q.4

संबंधित प्रश्न

Evaluate : `int(x-3)sqrt(x^2+3x-18)  dx`


Integrate the functions:

cot x log sin x


\[\int\sqrt{x - x^2} dx\]

Write a value of\[\int\frac{1}{1 + 2 e^x} \text{ dx }\].


Integrate the following functions w.r.t. x : `(x^2 + 2)/((x^2 + 1)).a^(x + tan^-1x)`


Integrate the following function w.r.t. x:

`(10x^9 +10^x.log10)/(10^x + x^10)`


Integrate the following functions w.r.t. x : `(2x + 1)sqrt(x + 2)`


Integrate the following functions w.r.t. x : tan5x


Integrate the following functions w.r.t. x :  tan 3x tan 2x tan x


Evaluate the following integrals : `int sqrt((e^(3x) - e^(2x))/(e^x + 1)).dx`


`int sqrt(1 + "x"^2) "dx"` =


Choose the correct alternative from the following.

`int "dx"/(("x" - "x"^2))`= 


Evaluate: `int 1/(2"x" + 3"x" log"x")` dx


Evaluate: `int "e"^sqrt"x"` dx


Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx


`int (sin  (5x)/2)/(sin  x/2)dx` is equal to ______. (where C is a constant of integration).


`int(1 - x)^(-2)` dx = `(1 - x)^(-1) + c`


Evaluate `int_(logsqrt(2))^(logsqrt(3)) 1/((e^x + e^-x)(e^x - e^-x)) dx`.


Evaluate 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 the following.

`int 1/(x^2 + 4x - 5)dx`


Evaluate `int 1/(x(x-1))dx`


Evaluate the following.

`int (x^3)/(sqrt(1 + x^4)) dx`


Evaluate `int(1+x+x^2/(2!))dx`


If f '(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).


Evaluate `int (1 + x + x^2/(2!)) dx`


Evaluate `int1/(x(x - 1))dx`


Evaluate the following.

`int1/(x^2 + 4x-5)dx`


`int (x + 1)/(x(1 + xe^x)) dx` is equal to


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×