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How do you integrate #x(x-1)^6 dx#?

Answer 1

Samantha Adkison

#int x(x-1)^6 dx = 1/8(x-1)^8+1/7(x-1)^7+C#

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#x(x-1)^6 = (x-1+1)(x-1)^6 = (x-1)^7 + (x-1)^6#

So:

#int x(x-1)^6 dx = 1/8(x-1)^8+1/7(x-1)^7+C#

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Answer 2

Luke Alvarez

To integrate ( x(x-1)^6 , dx ), you can use integration by parts method. Let ( u = x ) and ( dv = (x-1)^6 , dx ). Then, differentiate ( u ) to get ( du ), and integrate ( dv ) to get ( v ).

( u = x )
( du = dx )
( dv = (x-1)^6 , dx )
( v = \frac{1}{7}(x-1)^7 )

Now, apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

[ \int x(x-1)^6 , dx = \frac{x}{7}(x-1)^7 - \int \frac{1}{7}(x-1)^7 , dx ]

Now integrate the second term on the right-hand side using a substitution:

Let ( w = x - 1 )
( dw = dx )

[ \int \frac{1}{7}(x-1)^7 , dx = \frac{1}{7} \int w^7 , dw ]

[ = \frac{1}{7} \cdot \frac{1}{8} w^8 + C ]

[ = \frac{1}{56} (x-1)^8 + C ]

Finally, substitute this result back into the integration by parts formula:

[ \int x(x-1)^6 , dx = \frac{x}{7}(x-1)^7 - \frac{1}{56} (x-1)^8 + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.