Integration by Parts?

Question

Integral from 1 to 7=sqroot(t)lnt dt

Answer 1

say u =ln t
du = dt/t
dv = sqroot(t)
v = 2/3*t at power 3/2
Integration = uv - integration (v) du
After performing the integration insert the limits

Answer 2

Since sqrt(t) = t^(1/2) we have to compute Int t^(1/2) ln(t) dt. Put u = ln(t) and dv = t^(1/2) dt. Then Int udv = uv - Int v du

v = 2/3 t^(3/2) du = dt/t Therefore,

Int t^(1/2) ln(t) dt = 2/3 ln(t) t^(3/2) - Int 2/3 (t^3/2) dt/t = 2/3 ln(t) t^(3/2) - Int 2/3 t^(1/2) dt = 2/3 ln(t) t^(3/2) -2/3 * 2/3 (t^3/2) + C = 2/3 t^(3/2) [ln(t) - 2/3] + C Cis the integration constant

Answer 3

The basic formula is ∫udv = uv - ∫vdu
Let u= ln t and v = ∫√t

So ∫√t * ln t dt = ln t (2/3√t^3) – 2/3*∫√t^3 * t^-1 dt
= ln t (2/3√t^3) – 4/9*√t^3 +c
= √t^3 (2/3 ln t -4/9) + c
Evaluate between limits to get 16.24

Answer 4

u = ln(t)
du = dt / t

dv = t^(1/2) dt
v = (2/3) t^3/2

So you get
(2/3) t^3/2 * ln(t) - int.(2/3) t^3/2 * dt / t
= (2/3) t^3/2 * ln(t) - (2/3)*int. t^1/2 dt
= (2/3) t^3/2 * ln(t) - (4/9)*t^3/2
=(2/3) t^3/2 * [ ln(t) - 2/3]

Now substitute the limits to end up with
16.239

Answer 5

Integration by parts is caluculated as int(u*dv) = uv - int(v*du)

where int(...) is an integral

So,

let u = sqrt(t)
let dv = lnt

So u*v = sqrt(t)*(1/t)

- int(v*du) = -int((1/t)*(1/2)t^(-1/2)) = (-1/2)int((1/t)*t^(-1/2))

Now, 1/t is the same as t^-1 so, you get (-1/2)int(t^(-3/2)) =

(-1/2)*(-2)t^(-1/2) = t^(-1/2)

So the whole thing is sqrt(t)/t +(1/sqrt(t)).

Adding together you get 2/sqrt(t)

So, at 7, you get 2/sqrt(7)

at 1, you get 2/sqrt(1) = 2

So, the whole thing is (2/sqrt(7)) - 2 or 2*((1/sqrt(7) - 1)