can someone help me sove this ??????
Differentiate (x^2 + 1)^8;
Hence Find Sx(x^2 + 1)^7 dx
thanks all ... ps can you post it step by step so i can understand it as well
(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.