Question #cc7fa

2 Answers
mason m
Jun 21, 2017

#inte^root(3)(x+5)dx=3e^root(3)(x+5)((x+5)^(2/3)-2root(3)(x+5)+2)+C#

Explanation:

#inte^((x+5)^(1/3))dx#

Let #t=(x+5)^(1/3)#. This implies that #dt=1/3(x+5)^(-2/3)dx#.

This can be rearranged as #dx=3(x+5)^(2/3)dt=3t^2dt#. The integral then becomes:

#=inte^t(3t^2)dt#

Which we can perform integration by parts on:

#{(u=3t^2,=>,du=6tdt),(dv=e^tdt,=>,v=e^t):}#

#=3t^2e^t-int6te^tdt#

Again:

#{(u=6t,=>,du=6dt),(dv=e^tdt,=>,v=e^t):}#

#=3t^2e^t-(6te^t-int6e^tdt)#

#=3t^2e^t-6te^t+6e^t#

#=3e^t(t^2-2t+2)#

#=3e^root(3)(x+5)((x+5)^(2/3)-2root(3)(x+5)+2)+C#

Ratnaker Mehta
Jun 21, 2017

# 3{(x+5)^(2/3)-2(x+5)^(1/3)+2}e^(root3(x+5))+C.#

Explanation:

Let, #I=inte^(root3(x+5))dx.#

We subst., #x+5=t^3 rArr dx=3t^2dt.#

#:. I=int(e^t)(3t^2)dt.#

We will use the following Rule of Integration by Parts (IBP) :

#" IBP : "intuvdt=uintvdt-int{(du)/dt*intvdt}dt.#

We take, #u=3t^2 rArr (du)/dt=6t; v=e^t rArr intvdt=e^t.#

#:. I=3t^2e^t-int{(6t)(e^t)}dt=3t^2e^t-6intte^tdt, or,#

# I=3t^2e^t-6J, where, J=intte^tdt............(1).#

For #J,"we use IBP with "u=t, &, v=e^t.#

#:. J=tinte^tdt-int{(d/dtt)(inte^tdt)}dt,#

# J=te^t-inte^tdt=te^t-e^t..............(2).#

Combining #(1) and (2),# we have,

#I=3t^2e^t-6(te^t-e^t)=3t^2e^t-6te^t+6e^t,#

#=3(t^2-2t+2)e^t,# and, since, #t=(x+5)^(1/3),#

# I=3{(x+5)^(2/3)-2(x+5)^(1/3)+2}e^(root3(x+5))+C.#

Enjoy Maths.!

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