If you take u = x + 1
Then, x + 1 = u => sqrt(x+1) = sqrt(u) = u^(1/2)
Then, x = (u - 1)
And, du/dx = 1 => du = dx
So,
x*sqrt(x+1)*dx
Becomes
(u-1)*sqrt(u)*du = (u-1)u^(1/2)du
With the substitution u = x + 1
Find the substitution u to transform the integral into integral(u-1)u^(1/2)du. u=?
Then, x + 1 = u => sqrt(x+1) = sqrt(u) = u^(1/2)
Then, x = (u - 1)
And, du/dx = 1 => du = dx
So,
x*sqrt(x+1)*dx
Becomes
(u-1)*sqrt(u)*du = (u-1)u^(1/2)du
With the substitution u = x + 1
Let's try to find a substitution of the form u = g(x), such that when we substitute x with g(x), it simplifies the expression x+1. In this case, we want to eliminate the square root, so we need to make sure that after we make the substitution, the new expression no longer contains the square root.
We can try letting u = x + 1. By substituting x + 1 into u, we have u = x + 1.
To find du, we can differentiate both sides of the equation u = x + 1 with respect to x:
du/dx = d(x + 1)/dx
du/dx = 1
Rearranging, we find that du = dx.
Now, we can rewrite the integral using the substitution u = x + 1 and du = dx:
∫ xsqrt(x+1) dx = ∫ (u-1)u^(1/2) du.
Therefore, the substitution that transforms the integral into ∫ (u-1)u^(1/2) du is u = x + 1.