#\intx(2x+5)^8dx#?

so far I've worked out the u-sub to be #u=2x+5#, so #du=2dx\rArrdx=(du)/2#

3 Answers
Monzur R.
Sep 16, 2017

#intx(2x+5)^8dx = 1/360(18x-5)(2x+5)^9#

Explanation:

Your substitution is good, all you need to do is complete it and then integrate term by term.

#intx(2x+5)^8dx = 1/2int1/2(u-5)u^8du = 1/4intu^9-5u^8du = 1/4(1/10u^10 -5/9u^9)+"c" = 1/360(9(2x+5)^10-50(2x+5)^9)) +"c"=1/360(18x-5)(2x+5)^9#

Cesareo R.
Sep 16, 2017

See below.

Explanation:

#u=2x+8#, so #du=2dx\rArrdx=(du)/2#

#\intx(2x+8)^8dx equiv int (u-8)/2u^8 (du)/2 = 1/4intu^9 du-8/4int u^8 du = 1/40u^10-8/36u^9#

and now

#\intx(2x+8)^8dx = 1/40(2x+8)^10-8/36(2x+8)^9+C#

Ratnaker Mehta
Sep 17, 2017

# 1/360(2x+5)^9(18x-5)+C.#

Explanation:

It is confirmed from the Questioner that, the Problem is,

#intx(2x+5)^8dx.#

Let us solve the Problem without using the Method of Substitution.

Note that,

#int(ax+b)^ndx=1/a*(ax+b)^(n+1)/(n+1)+c, ane0, n!=-1...(star)#

Suppose that, #I=intx(2x+5)^8dx,#

#:. I=1/2int(2x)(2x+5)^8dx,#

#=1/2int{(2x+5)-5}(2x+5)^8dx,#

#=1/2int{(2x+5)(2x+5)^8-5(2x+5)^8}dx,#

#=1/2int{(2x+5)^9-5(2x+5)^8}dx,#

#=1/2int(2x+5)^9dx-5/2int(2x+5)^8dx.#

Using #(star),# now, we get,

#I=1/2*1/2(2x+5)^(9+1)/(9+1)-5/2*1/2(2x+5)^(8+1)/(8+1),#

#=1/40(2x+5)^10-5/36(2x+5)^9.#

# rArr I=1/360(2x+5)^9(18x-5)+C.#

Enjoy Maths.!

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