∫5ex(ex+1)(e2x+9) dx - Mathematics and Statistics

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Sum

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

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Solution

Let i = `int (5"e"^x)/(("e"^x + 1)("e"^(2x) + 9))  "d"x`

Put ex = t

∴ ex dx = dt

∴ I = `int  (5"dt")/(("t" + 1)("t"^2 + 9))`

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) + 2"B" + 2(1/2)`

∴ – 1 = 2B

∴ B = `-1/2`

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

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

= `1/2[log|"t" + "t"| - 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"`

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

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