# Math Help - Integration By Parts

1. ## Integration By Parts

Let INT be the integral sign.

Use integration by parts to integrate

INT t*ln(t+1) dt

(1) Do I let u = In(t+1) and dv = tdt?

(2) Must I integrate by parts twice in this sample?

(3) Can someone get me started?

2. ## Re: Integration By Parts

why not try both and see what you learn

3. ## Re: Integration By Parts

I decided to let u = ln(t+1) and
dv = txt. I then found v to be (t^2)/2
and du = dt/(t+1).

I used uv-INT vdu

My answer after working out the math is the following:

(1/2)[t^2*(t+1) - (t^3)/3 + ln(t+1)] + C

Right?

4. ## Re: Integration By Parts

Originally Posted by nycmath
I decided to let u = ln(t+1) and
dv = txt. I then found v to be (t^2)/2
and du = dt/(t+1).

I used uv-INT vdu

My answer after working out the math is the following:

(1/2)[t^2*(t+1) - (t^3)/3 + ln(t+1)] + C

Right?
it doesn't appear to be. The answer is

Spoiler:
$\Large -\frac{t^2}{4}+\frac{1}{2} t^2 \log (t+1)+\frac{t}{2}-\frac{1}{2} \log (t+1)$

5. ## Re: Integration By Parts

Is my selection for u and dv correct?

Is my du and v correct?

Must I apply integration by parts twice?

Can you get me started?

I will do the rest....

6. ## Re: Integration By Parts

Is my selection for u and dv correct?

Is my du and v correct?

Must I apply integration by parts twice?

Can you get me started?

I will do the rest....

7. ## Re: Integration By Parts

Originally Posted by nycmath
Is my selection for u and dv correct?

Is my du and v correct?

Must I apply integration by parts twice?

Can you get me started?

I will do the rest....
there is no "correct" vs. "incorrect", both will work. It's just a matter of which turns out to be easier. I suspect you are doing it correctly you just got some algebra or something wrong. If you want post precisely what you have done and I'll see if I can figure out where you went wrong.

8. ## Re: Integration By Parts

Originally Posted by nycmath
I decided to let u = ln(t+1) and
dv = txt. I then found v to be (t^2)/2
and du = dt/(t+1).

I used uv-INT vdu

My answer after working out the math is the following:

(1/2)[t^2*(t+1) - (t^3)/3 + ln(t+1)] + C

Right?
Assuming that you meant \displaystyle \begin{align*} dv = t\,dt \end{align*}, I agree with your setting of the functions, so you should get

\displaystyle \begin{align*} \int{ t\ln{(t + 1)} \, dt} &= \frac{t^2}{2}\ln{(t + 1 )} - \int{\frac{t^2\,dt}{t + 1}} \\ &= \frac{t^2}{2}\ln{(t + 1)} - \int{ \frac{ \left( u - 1 \right) ^2 \, du }{u} } \textrm{ after letting } u = t + 1 \implies du = dt \\ &= \frac{t^2}{2} \ln{(t + 1)} - \int{ \frac{ \left( u^2 - 2u + 1 \right) \, du}{u} } \\ &= \frac{t^2}{2}\ln{(t + 1)} - \int{ u - 2 + \frac{1}{u} \, du } \\ &= \frac{t^2}{2}\ln{(t + 1)} - \left( \frac{u^2}{2} - 2u + \ln{|u|} \right) + C \\ &= \frac{t^2}{2} \ln{(t + 1)} - \frac{ \left( t + 1 \right) ^2}{2} + 2 \left( t + 1 \right) - \ln{ \left| t + 1 \right| } + C \end{align*}

9. ## Re: Integration By Parts

Nice work, Prove It. I will try a similar problem on Monday and post my complete reply. Enjoy your weekend.

10. ## Re: Integration By Parts

[QUOTE=romsek;818121]there is no "correct" vs. "incorrect", both will work. It's just a matter of which turns out to be easier. I suspect you are doing it correctly you just got some algebra or something wrong. If you want post precisely what you have done and I'll see if I can figure out where you went wron

I will try a similar problem on Monday and post my complete reply. Enjoy your weekend.

11. ## Re: Integration By Parts

[QUOTE=nycmath;818127]
Originally Posted by romsek
there is no "correct" vs. "incorrect", both will work. It's just a matter of which turns out to be easier. I suspect you are doing it correctly you just got some algebra or something wrong. If you want post precisely what you have done and I'll see if I can figure out where you went wron

I will try a similar problem on Monday and post my complete reply. Enjoy your weekend.
for my edification I wrote up a quick sheet to verify that both methods work

The first line computes the integral so we know what the answer is.

then we integrate by parts with u=t, etc.

At the end of the first bit we take the difference of the integral and the integral found by integrating by parts and see that it is 0.

In the second part we flip u and dv and show the same thing.

12. ## Re: Integration By Parts

[QUOTE=romsek;818131]
Originally Posted by nycmath

for my edification I wrote up a quick sheet to verify that both methods work

The first line computes the integral so we know what the answer is.

then we integrate by parts with u=t, etc.

At the end of the first bit we take the difference of the integral and the integral found by integrating by parts and see that it is 0.

In the second part we flip u and dv and show the same thing.
Before I go to bed in a few minutes, can you show me the other method letting u = t and dv everything else using latex? I will check your answer in the morning. Thanks.

13. ## Re: Integration By Parts

Thank you romsek and Prove It. Good night. More integration by parts on Monday.

14. ## Re: Integration By Parts

Originally Posted by Prove It
Assuming that you meant \displaystyle \begin{align*} dv = t\,dt \end{align*}, I agree with your setting of the functions, so you should get

\displaystyle \begin{align*} \int{ t\ln{(t + 1)} \, dt} &= \frac{t^2}{2}\ln{(t + 1 )} - \int{\frac{t^2\,dt}{t + 1}} \\ &= \frac{t^2}{2}\ln{(t + 1)} - \int{ \frac{ \left( u - 1 \right) ^2 \, du }{u} } \textrm{ after letting } u = t + 1 \implies du = dt \\ &= \frac{t^2}{2} \ln{(t + 1)} - \int{ \frac{ \left( u^2 - 2u + 1 \right) \, du}{u} } \\ &= \frac{t^2}{2}\ln{(t + 1)} - \int{ u - 2 + \frac{1}{u} \, du } \\ &= \frac{t^2}{2}\ln{(t + 1)} - \left( \frac{u^2}{2} - 2u + \ln{|u|} \right) + C \\ &= \frac{t^2}{2} \ln{(t + 1)} - \frac{ \left( t + 1 \right) ^2}{2} + 2 \left( t + 1 \right) - \ln{ \left| t + 1 \right| } + C \end{align*}
Hello. Where did (u-1)^2 in your second step come from?

15. ## Re: Integration By Parts

Originally Posted by nycmath
Hello. Where did (u-1)^2 in your second step come from?
Since he substituted u = t + 1 then t = u - 1

------

Tip for you :
You should know that you need to let u = ln(t+1)
because if you let dv = ln(t+1) dt then it needs integration by parts to get v itself

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