Integrate t^(1/2)ln(t) dt from x=8 to x=1.?

Question

Integrate t^(1/2)ln(t) dt from x=8 to x=1.?

Answer 1

∫ t^(1/2) ln(t) dt =

let t^(1/2) = u →

t = u² →

dt = 2u du

thus, substituting, you get:

∫ t^(1/2) ln(t) dt = ∫ u ln(u²) 2u du =

2 ∫ u² ln(u²) du =

now, integrating it by parts, let:

ln(u²) = U → (2u/u²) du = (2/u) du = dU

u² du = v → [u^(2+1)]/(2+1) = (1/3) u³ = v

thus, integrating:

∫ U dv = U v - ∫ v dU →

2 ∫ u² ln(u²) du = 2 [(1/3) u³ ln(u²) - ∫ (1/3) u³ (2/u) du] =

2 [(1/3) u³ ln(u²) - (2/3) ∫ u² du] =

(2/3) u³ ln(u²) - (4/3) ∫ u² du =

(2/3) u³ ln(u²) - (4/3) [u^(2+1)]/(2+1) + C =

(2/3) u³ ln(u²) - (4/3)(1/3) u³ + C =

(2/3) u³ ln(u²) - (4/9) u³ + C

thus, substituting back u = t^(1/2), the antiderivative is:

(2/3) t^(3/2) ln(t) - (4/9) t^(3/2) + C

finally, evaluate the definite integral from 8 to 1 (assuming that "1." means 1) :

[(2/3) 1^(3/2) ln(1) - (4/9)1^(3/2)] - [(2/3) 8^(3/2) ln(8) - (4/9) 8^(3/2)] =

[0 - (4/9)] - [(2/3) 8^(3/2) ln(8) - (4/9) 8^(3/2)] =

- (4/9) - (2/3) 8^(3/2) ln(2³) + (4/9) 8^(3/2) =

(applying log properties)

- (4/9) - (2/3) (8) 8^(1/2) 3 ln(2) + (4/9) (8) 8^(1/2) =

(canceling 3)

- (4/9) - 2 (8) (2) 2^(1/2) ln(2) + (4/9) (8)(2) 2^(1/2) =

- (4/9) - (32) 2^(1/2) ln(2) + (64/9) 2^(1/2)

I hope it helps...

Bye!