∫dx/((x+1)√x)
Thanks a lot!!!!!!
Thanks a lot!!!!!!
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∫(1 to infinity, 1/((x + 1)√x) dx )
Like all improper integrals, you must change this into a limit.
lim ∫(1 to t, 1/((x + 1)√x) dx )
t -> infinity
Use u-substitution.
Let u = √x. That means
u^2 = x; therefore
2u du = dx
When x = 1, u = √1 = 1.
When x = t, u = √t. Therefore, our new bounds are from 1 to √t.
lim ∫(1 to √t, 1/((u^2 + 1)u) 2u du )
t -> infinity
lim ∫(1 to √t, 1/((u^2 + 1)) 2 du )
t -> infinity
2 * lim ∫(1 to √t, 1/((u^2 + 1)) du )
t -> infinity
This is a known derivative; it is the derivative of arctan.
2 * lim arctan(u) (evaluated from 1 to √t)
t -> infinity
2 * lim [arctan(√t) - arctan(1)]
t -> infinity
And as t approaches infinity, arctan approaches pi/2.
Additionally, arctan(1) is equal to pi/4.
2 * [ pi/2 - pi/4 ]
pi - pi/2
which is just
pi/2
Like all improper integrals, you must change this into a limit.
lim ∫(1 to t, 1/((x + 1)√x) dx )
t -> infinity
Use u-substitution.
Let u = √x. That means
u^2 = x; therefore
2u du = dx
When x = 1, u = √1 = 1.
When x = t, u = √t. Therefore, our new bounds are from 1 to √t.
lim ∫(1 to √t, 1/((u^2 + 1)u) 2u du )
t -> infinity
lim ∫(1 to √t, 1/((u^2 + 1)) 2 du )
t -> infinity
2 * lim ∫(1 to √t, 1/((u^2 + 1)) du )
t -> infinity
This is a known derivative; it is the derivative of arctan.
2 * lim arctan(u) (evaluated from 1 to √t)
t -> infinity
2 * lim [arctan(√t) - arctan(1)]
t -> infinity
And as t approaches infinity, arctan approaches pi/2.
Additionally, arctan(1) is equal to pi/4.
2 * [ pi/2 - pi/4 ]
pi - pi/2
which is just
pi/2