Evaluate the integral using integration by parts where possible. (t2 − t)ln(−t) dt

help please

Hello,

let's write the integral as:

∫ [ln (- t)] (t² - t) dt =

let:

ln (- t) = u → (differentiating) → (- 1) [1 /(- t)] dt = (1 /t) dt = du

(t² - t) dt = dv → (integrating) → [1/(2+1)] t^(2+1) - [1/(1+1)] t^(1+1) = (1/3)t³ - (1/2)t² = v

then, integrating by parts:

∫ u dv = v u - ∫ v du

∫ [ln (- t)] (t² - t) dt = [(1/3)t³ - (1/2)t²] ln (- t) - ∫ [(1/3)t³ - (1/2)t²] (1 /t) dt =

[(1/3)t³ - (1/2)t²] ln (- t) - ∫ [(1/3)t³ (1 /t) - (1/2)t² (1 /t)] dt =

(simplifying)

[(1/3)t³ - (1/2)t²] ln (- t) - ∫ [(1/3)t² - (1/2)t] dt =

(splitting into two integrals and pulling constants out)

[(1/3)t³ - (1/2)t²] ln (- t) - (1/3) ∫ t² dt - (-1/2) ∫ t dt =

[(1/3)t³ - (1/2)t²] ln (- t) - (1/3) [1/(2+1)] t^(2+1) + (1/2) ∫ t dt =

[(1/3)t³ - (1/2)t²] ln (- t) - (1/3)(1/3)t³ + (1/2) [1/(1+1)] t^(1+1) + C =

[(1/3)t³ - (1/2)t²] ln (- t) - (1/9)t³ + (1/2)(1/2)t² + C =

ending with:


[(1/3)t³ - (1/2)t²] ln (- t) - (1/9)t³ + (1/4)t² + C



I hope it helps