Question

Integrate the following with respect to x:

x⁵e^x2

Answer 1

`int x^5 "e"^(x^2) "d"x = int (x2)^2 * "e"^(x^2) * x  "d"x`

Put x² = t

2x dx = dt

⇒ x dx = `1/2  "dt"`

`int x^5 "e"^(x^2)  "d"x = int "t"^2 * "e"^"t" * 1/2  "dt"`

= `1/2 int "t"^2 "e"^"t"  "dt"`  .........(1)

Consider `int "t"^2  "e"^"t"  "dt"`

u = t²

u' = 2t

u" = 2

u"' = 0

dv = e^t dt

⇒ v = `int "e"^"t"  "dt"`

⇒ v = e^t

v₁ = `int "v"  "dt"`

= `int "e"^"t" "dt"`

= e^t

v₂ = `int "v"_1  "dt"`

= `int "e"^"t" "dt"`

= e^t 

v₃ = `int "v"_2  "dt"`

= `int "e"^"t" "dt"`

= e^t 

`int u"  "dv"` = uv – u'v₁ + u"v₂ – u"'v₃ + u"'v₄ – ...........

`int "t"^2 "e"^"t"  "dt"` = t² e^t – 2t e^t + 2e^t + 0e^t

= `("t"^2 - 2"t" + 2) "e"^"t" + "c"`

Substtuting in equation (1) we get

`int x^5  "e"^(x^2)  "d"x = 1/2["t"^2 - 2"t" + 2] "e"^"t" + "c"`

= `1/2 [(x^2)^2 - 2x^2 + 2] "e"^(x^2) + "c"`

`int x^5  "e"^(x^2)  "d"x =  1/2 [x^4 - 2x^2 + 2] "e"^(x^2) + "c"`