Find the integral. (Note: Solve by the simplest method )

[chapsticks]

Homework Statement

Find the integral. (Note: Solve by the simplest method - not all require integration by parts.)
∫t ln(t + 2) dt

Homework Equations

∫udv=uv-∫vdu

The Attempt at a Solution

dv=tdt v=tdt=t²/2
u=ln(t+2) du=1/(t+2)dt

∫t ln(t + 2) dt
=(t²/2)ln(t+2)-(1/2)∫(t²/2)dt
=(t²/2)ln(t+2)-(1/2)∫(t-2+(4/(t+2))dt
=(t²/2)ln(t+2)-(1/2)[(t²/2)-2t+4ln(t+2)]+C
=? Idk why I keep getting it wrong?

 

[LCKurtz]

Homework Statement

Find the integral. (Note: Solve by the simplest method - not all require integration by parts.)
∫t ln(t + 2) dt

Homework Equations

∫udv=uv-∫vdu

The Attempt at a Solution

dv=tdt v=tdt=t²/2
u=ln(t+2) du=1/(t+2)dt

Leave that red part out.

∫t ln(t + 2) dt
=(t²/2)ln(t+2)-(1/2)∫(t²/2)dt

Check the vdu in that last integral.

 

[SammyS]

Homework Statement

Find the integral. (Note: Solve by the simplest method - not all require integration by parts.)
∫t ln(t + 2) dt

Homework Equations

∫udv=uv-∫vdu

The Attempt at a Solution

dv=tdt v=tdt=t²/2
u=ln(t+2) du=1/(t+2)dt

∫t ln(t + 2) dt
=(t²/2)ln(t+2)-(1/2)∫(t²/2)dt

The above line should be:

$\displaystyle (t^2/2)\ln(t+2)-(1/2)\int \frac{t^2}{t+2}\,dt\,.$​

=(t²/2)ln(t+2)-(1/2)∫(t-2+(4/(t+2))dt
=(t²/2)ln(t+2)-(1/2)[(t²/2)-2t+4ln(t+2)]+C
=? Idk why I keep getting it wrong?

 

[xaos]

it reads okay to me, although you missed a step. are you not simplifying enough?