intx(x+1)^3dx=?

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2 Answers
Apr 14, 2018

therefore " Option A"

Explanation:

Approached by u substitution...

Let

u = x+1

u-1 = x

=> du = dx

=> int (u-1)u^3 du

=> int u^4 - u^3 du

=> 1/5 u^5 - 1/4 u^4 + c

=> 1/5 (x+1)^5 - 1/4 (x+1)^4 + c

therefore " Option A"

Apr 14, 2018

The answer is "option (A)"

Explanation:

Let u=x+1, =>, du=dx

x=u-1

Therefore,

intx(x+1)^3=int(u-1)u^3du

=int(u^4-u^3)du

=u^5/5-u^4/4

=(x+1)^5/5-(x+1)^4/4+C

=(x+1)^4/20((4x+4-5))+C

=(x+1)^4/20(4x-1)+C

The answer is "option (A)"