Calculus substitution question

[Scarab of the East]

Here's the problem:

x(x-1)^5 dx

I have no clue how to proceed. I was knocking out the easier substitution problems with no sweat, but apparently I don't fully grasp it because I am looking at the answer and still don't see it.

 

[Scarab of the East]

I think the problem I'm having is finding du. Here's what I'm getting (not what the book has):

x(x-1)^5 dx

u= x-1
du= dx

1/6(x-1)^6+C as the final answer

Where does the x outside the inside function go? I know that if it was a constant k there, it would be du= kdx, but as far as I can understand what the book is saying, du should be the derivative of u, so in this problem it would be 1...

 

[Psyduck]

rewrite as

Int (x - 1 + 1) (x-1)^5 dx = Int (x-1)^6 dx + Int (x-1)^5 dx

 

[Captain.Heroin]

Frankly I can't make heads or tails of this. Are you actually planning on using calculus in the real world? I knew I would never be doing anything like that and gave up on learning more math deliberately because of this.

I think it's a waste of people's time and energy to learn useless information. When I say useless information, it's not that the information is inherently useless, but that your life may not present any worthwhile, useful attempts to utilize this information, making it by default useless.

 

[Dead_Flowers]

A little late, but...

Let u=x-1. Then, du=dx and x=u+1.

Rewriting the integral, we have: int[(u+1)u^5 du]= int[u^6 + u^5 du] = (1/7)u^7 + (1/6)u^6 + C

Finally, substiute x-1 for u, and the integral is (1/7)(x-1)^7 +(1/6)(x-1)^6 + C.

As for this being a waste of time, many fields will use these exact techniques quite often. Even if it is not utilized in one's career, this is beautiful stuff and will certainly enhance the student's quantitative and critical thinking skills.

 

[Jerry Atrick]

A little late, but...

Next time, try to post your answer before the OP's homework is due