∫e3x-e2xex+1 dx - Mathematics and Statistics

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Sum

`int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1))  "d"x`

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Solution

Let I = `int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1))  "d"x`

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

= `int"e"^x sqrt(("e"^x - 1)/("e"^x + 1))  "d"x`

Put ex = t

∴ ex dx = dt

∴ I = `int sqrt(("t" - 1)/("t" + 1))  "dt"`

= `int sqrt(("t" - 1)/("t" + 1) xx ("t" - 1)/("t" - 1))  "dt"`

= `int ("t" - 1)/sqrt("t"^2 - 1)  "dt"`

= `int ("t"/sqrt("t"^2 - 1) - 1/sqrt("t"^2 - 1))  "dt"`

= `int "t"/sqrt("t"^2 - 1)  "dt" - int  1/sqrt("t"^2 - 1)  "dt"`

= I1 − I2      .......(i)

I1 = `int "t"/sqrt("t"^2 - 1)  "dt"`

Put t2 − 1 = a

∴ 2t dt = da

I1 = `1/2 int "da"/sqrt("a")`

= `1/2 int "a"^(1/2) "da"`

= `1/2("a"^(1/2)/(1/2)) + "c"_1`

= `sqrt("a") + "c"_1`

= `sqrt("t"^2 - 1) + "c"_1`

I1 = `sqrt("e"^(2x) - 1) + "c"_1`    ......(ii)

I2 = `int 1/sqrt("t"^2 - 1^2)  "dt"`

= `log|"t" + sqrt("t"^2 - 1^2)| + "c"_2`

I2 = `log|"e"^x + sqrt("e"^(2x) - 1)| + "c"_2`  .......(iiii)

 From (i), (ii) and (iii), we get

I = `sqrt("e"^(2x) - 1) - log|"e"^x + sqrt("e"^(2x) - 1)| +"c"`,

 where c = c1 − c2

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

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