(x^2 +1)^8

You know d/dx (u)^n

= n*[(u)^(n-1)]*du ?

Let y = (x^2 +1)^8

Differentiate both sides with respect to x,

dy/dx = 8*(x^2 +1)^7 *(2x +0)

dy/dx = 8(x^2 +1)^7 *(2x)

dy/dx = 16x(x^2 +1)^7 ----****

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Sx(x^2 + 1)^7 dx

INT.[x(x^2 +1)^7]dx

You also know

INT.[u^n]du = 1/(n+1) *[u^(n+1)] ?

Be sure that the "du" outside the bracket is really the "du" of the "u" inside the bracket.

INT.[x(x^2 +1)^7]dx

The dx, or 1, outside the bracket is not the d/dx of x(x^2 +1)^7 inside the bracket.

What if we bring out the x?

= INT.[(x^2 +1)^7](x dx)

The (x) is not the d/dx of the (x^2 +1).

What is the d/dx of (x^2 +1)?

It is 2x.

So if we make the (x dx) outside the bracket into (2x dx), then this (2x) is already the d/dx of (x^2 +1).

So we can integrate now?

Not yet.

In making (x dx) into (2x dx) we multiplied the (x dx) by 2. So we have to divide by 2, or multiply by (1/2), the new (2x dx) so that the original (x dx) is maintained.

Hence,

= INT.[(x^2 +1)^7](2x/2 dx)

= INT.[(x^2 +1)^7](2x dx)(1/2)

= (1/2)INT.[(x^2 +1)^7](2x dx)

Now we can integrate,

= (1/2)*(1/8)(x^2 +1)^8 +C

= (1/16)(x^2 +1)^8 +C ----answer.

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You can ask if you want more explanations re this answer, as I do not know your "strength" in Calculus.