English

Evaluate: ∫5ex(ex+1)(e2x+9)dx

Advertisements
Advertisements

Question

Evaluate:

`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`

Evaluate
Advertisements

Solution

Let I = `int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`

Put ex = t

⇒ exdx = dt

∴ I = `int  5/((t + 1)(t^2 + 9))dt`

Let `5/((t + 1)(t^2 + 9))`

= `A/(t + 1) + (Bt + C)/(t^2 + 9)`

∴ 5 = A(t2 + 9) + (Bt + C)(t + 1)    ...(i)

Putting t = –1 in (i), we get

5 = A[(–1)2 + 9]

∴ 5 = 10A

∴ A = `1/2`

Putting t = 0 in (i), we get

5 = A(0 + 9) + (0 + C) (0 + 1)

∴ 5 = 9A + C

∴ 5 = `9(1/2) + "C"`

∴ C = `1/2`

Putting t = 1 in (i), we get

5 = A(12 + 9) + (B + C)(1 + 1)

∴ 5 = 10A + 2B + 2C

∴ 5 = `10(1/2) + 2B + 2(1/2)`

∴ – 1 = 2B

∴ B = `-1/2`

∴ `5/((t + 1)(t^2 + 9)) = (1/2)/(t + 1) + (1/2t + 1/2)/(t^2 + 9)`

∴ I = `int((1/2)/(t+ 1) + ((-1)/2t + 1/2)/(t^2 + 9)) dt`

= `1/2 [int 1/(t + 1) dt - int  t/(t^2 + 9) dt + int  1/(t^2 + 9) dt]`

= `1/2[int 1/(t + 1) dt - 1/2  int (2t)/(t^2 + 9) dt + int 1/(t^2 + 3^2) dt]`

= `1/2 [log (t + 1) - 1/2 log (t^2 + 9) + 1/3tan^-1(t/3)] + c`

∴ I = `1/2  log (e^x + 1) - 1/4  log (e^(2x) + 9) + 1/6  tan^-1 ((e^x)/3) + c`

shaalaa.com
  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Long Answers III

RELATED QUESTIONS

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


Evaluate: `∫8/((x+2)(x^2+4))dx` 


Integrate the rational function:

`1/(x^2 - 9)`


Integrate the rational function:

`x/((x^2+1)(x - 1))`


Integrate the rational function:

`x/((x -1)^2 (x+ 2))`


Integrate the rational function:

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


`int (dx)/(x(x^2 + 1))` equals:


Integrate the following w.r.t. x : `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3))`


Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`


Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`


Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`


Integrate the following w.r.t. x : `(1)/(x^3 - 1)`


Integrate the following w.r.t. x : `(1)/(sinx*(3 + 2cosx)`


Integrate the following w.r.t. x : `(5*e^x)/((e^x + 1)(e^(2x) + 9)`


Integrate the following w.r.t.x: `(x + 5)/(x^3 + 3x^2 - x - 3)`


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


`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`


`int 1/(x(x^3 - 1)) "d"x`


If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)


`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`


`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`


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


`int ("d"x)/(2 + 3tanx)`


`int xcos^3x  "d"x`


Evaluate `int x^2"e"^(4x)  "d"x`


`int 1/(4x^2 - 20x + 17)  "d"x`


`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5)  "dt"`


The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, the denominator becomes eight times the numerator. Find the fraction.


Find: `int x^2/((x^2 + 1)(3x^2 + 4))dx`


If `int 1/((x^2 + 4)(x^2 + 9))dx = A tan^-1  x/2 + B tan^-1(x/3) + C`, then A – B = ______.


If `intsqrt((x - 5)/(x - 7))dx = Asqrt(x^2 - 12x + 35) + log|x| - 6 + sqrt(x^2 - 12x + 35) + C|`, then A = ______.


Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`


Evaluate: 

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


Evaluate:

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


Evaluate.

`int (5x^2 - 6x + 3) / (2x -3) dx`


Evaluate:

`int(2x^3 - 1)/(x^4 + x)dx`


Which expression defines a rational function?


A proper rational function can be expressed as a sum of simpler rational functions called what?


Which partial form is appropriate for \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})^{2}(x-\mathrm{b})}\]?


What should be checked before beginning partial-fraction decomposition?


What is done after long division, before writing the appropriate partial-fraction decomposition?


How are the constants \[\mathrm{A},\mathrm{B},\mathrm{C},\ldots\] determined in a partial-fraction decomposition?


Evaluate \[\int\frac{x^{2}+1}{x^{2}-5x+6}\,dx\].


What numerator should be used for each distinct linear factor in a partial-fraction decomposition?


What must be included for a repeated linear factor?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×