Help with Integration by Parts

[BlBl]

I'm trying to do as many integration by parts problems as I can in order to get practice, but for some reason I keep getting stuck on some and just can't get anywhere with them (I move on to ones I can do though). For example here's a problem that I have the answer to but I just can't seem to imagine the steps that were taken to get the answer.

$\displaystyle \int (t+2)\sqrt{2+3t}$ or $\displaystyle \int (t+2)({2+3t}^{1/2})$

My preferred pattern for integration by parts goes:

$\displaystyle \int fg'=fg-\int f'g$

So applying this to the problem I get:

$\displaystyle \int (t+2)({2+3t}^{1/2})=(t+2)(\dfrac{2}{3}(2+3t)^{3/2}-\int (\dfrac{t^2}{2}+2t)(\dfrac{2}{3}(2+3t)^{3/2}$

Chegg/Cramster and Wolfram are confusing to follow as well. I don't know, maybe I'm just in a rut with some of these.

 

[soroban]

Hello, BlBl!

$\displaystyle \displaystyle I \;=\;\int (t+2)\sqrt{2+3t}\,dx$

My preferred pattern for integration by parts goes: .$\displaystyle \int fg'\:=\:fg-\int f'g$

So applying this to the problem I get:

$\displaystyle \displaystyle \int (t+2)({2+3t}^{1/2})\;=\;(t+2)\tfrac{2}{3}(2+3t)^{3/2}-\int (\tfrac{1}{2}t^2+2t)\tfrac{2}{3}(2+3t)^{3/2}$
You integrated both parts?
And you forgot the 3 in "3t".

I was taught a rule for choosing the parts:
. . Let $\displaystyle dv$ be the hardest part you can still integrate.

$\displaystyle \begin{Bmatrix}u &=& t + 2 && dv &=& (2 + 3t)^{\frac{1}{2}}dt \\ du &=& dt && v &=& \frac{2}{9}(2 + 3t)^{\frac{3}{2}} \end{Bmatrix}$


$\displaystyle \displaystyle\text{Hence: }\:I \;=\;\tfrac{2}{9}(t+2)(2+3t)^{\frac{3}{2}} - \tfrac{2}{9}\int(2+3t)^{\frac{3}{2}}dt$

Can you finish it now?

 

[Deleted member 4993]

I'm trying to do as many integration by parts problems as I can in order to get practice, but for some reason I keep getting stuck on some and just can't get anywhere with them (I move on to ones I can do though). For example here's a problem that I have the answer to but I just can't seem to imagine the steps that were taken to get the answer.

$\displaystyle \int (t+2)\sqrt{2+3t}$ or $\displaystyle \int (t+2)({2+3t}^{1/2})$

My preferred pattern for integration by parts goes:

$\displaystyle \int fg'=fg-\int f'g$

So applying this to the problem I get:

$\displaystyle \int (t+2)({2+3t}^{1/2})=(t+2)(\dfrac{2}{3}(2+3t)^{3/2}-\int (\dfrac{t^2}{2}+2t)(\dfrac{2}{3}(2+3t)^{3/2}$

Chegg/Cramster and Wolfram are confusing to follow as well. I don't know, maybe I'm just in a rut with some of these.

You need to be more careful about what you write - and review what you are writing.

$\displaystyle \int (t+2)\sqrt{2+3t} dt$

=$\displaystyle \int (t+2) * ({2+3t})^{\frac{1}{2}} dt$

f = t + 2 → f' = dt

$\displaystyle g' = ({2+3t})^{\frac{1}{2}} dt$ → $\displaystyle g = \frac{2}{3} * ({2+3t})^{\frac{3}{2} } * \frac{1}{3}$ → $\displaystyle g = \frac{2}{9} * ({2+3t})^{\frac{3}{2} }$

At this point, you should check your g by differentiating - and making sure you got back the original function.

Now continue.....

 

[BlBl]

Hello, BlBl!


I was taught a rule for choosing the parts:
. . Let $\displaystyle dv$ be the hardest part you can still integrate.

$\displaystyle \begin{Bmatrix}u &=& t + 2 && dv &=& (2 + 3t)^{\frac{1}{2}}dt \\ du &=& dt && v &=& \frac{2}{9}(2 + 3t)^{\frac{3}{2}} \end{Bmatrix}$


$\displaystyle \displaystyle\text{Hence: }\:I \;=\;\tfrac{2}{9}(t+2)(2+3t)^{\frac{3}{2}} - \tfrac{2}{9}\int(2+3t)^{\frac{3}{2}}dt$

Can you finish it now?

You need to be more careful about what you write - and review what you are writing.

$\displaystyle \int (t+2)\sqrt{2+3t} dt$

=$\displaystyle \int (t+2) * ({2+3t})^{\frac{1}{2}} dt$

f = t + 2 → f' = dt

$\displaystyle g' = ({2+3t})^{\frac{1}{2}} dt$ → $\displaystyle g = \frac{2}{3} * ({2+3t})^{\frac{3}{2} } * \frac{1}{3}$ → $\displaystyle g = \frac{2}{9} * ({2+3t})^{\frac{3}{2} }$

At this point, you should check your g by differentiating - and making sure you got back the original function.

Now continue.....

Yes, I see that I forgot the 1/3 factor that would come from the 3. Finishing off what y'all have started I get:

$\displaystyle \tfrac{2}{9}(t+2)(2+3t)^{\frac{3}{2}} - \tfrac{2}{9}\int(2+3t)^{\frac{3}{2}}dt=\tfrac{2}{9}(t+2)(2+3t)^{\frac{3}{2}} - \tfrac{2}{9}\cdot \tfrac{2}{5}(2+3t)^{\frac{5}{2}}dt$

=

$\displaystyle \int (t+2)\sqrt{2+3t} dt =\tfrac{2}{9}(t+2)(2+3t)^{\frac{3}{2}} - \tfrac{4}{45}(2+3t)^{\frac{5}{2}}dt$

But this still isn't right. Right? What am I missing at this point?

 

[Deleted member 4993]

You are forgetting about "3t" again - for your second integration.

 

[BlBl]

You are forgetting about "3t" again - for your second integration.

Yes, this is where I'm still lost. Should I multiply it all by $\displaystyle \dfrac{3t^2}{2}$? Should I try to attempt a second integration by parts?

I haven't done enough of these to feel comfortable that I know what I'm doing.

 

[Deleted member 4993]

Yes, this is where I'm still lost. Should I multiply it all by $\displaystyle \dfrac{3t^2}{2}$? Should I try to attempt a second integration by parts?

I haven't done enough of these to feel comfortable that I know what I'm doing.

No...

You need to review your simple integration skills - before you try these complex function.

Try it this way - just the second integration:

let u = 2 + 3t → du = 3 dt

then your second integration becomes:



$\displaystyle \tfrac{2}{9}\int(2+3t)^{\frac{3}{2}}dt$

= $\displaystyle \tfrac{2}{9}\int(u)^{\frac{3}{2}} \frac{du}{3}$

= $\displaystyle \tfrac{2}{27}\int(u)^{\frac{3}{2}} du$

= $\displaystyle \tfrac{2}{27} * \frac{2}{5}(u)^{\frac{5}{2}} + C$

= $\displaystyle \tfrac{4}{135} * (2 + 3t)^{\frac{5}{2}} + C$

do not skip steps .... you are not ready for it.

 

[BlBl]

Thank you very much Subhotosh

I agree. I'm not ready for it. But I want to be able to get there quickly. I thought that I'd done enough integrals to understand this stuff. It seems I've gotta do a lot more.

 

[b4nanas]

Integrate by substitution

no...

You need to review your simple integration skills - before you try these complex function.

Try it this way - just the second integration:

Let u = 2 + 3t → du = 3 dt

then your second integration becomes:



$\displaystyle \tfrac{2}{9}\int(2+3t)^{\frac{3}{2}}dt$

= $\displaystyle \tfrac{2}{9}\int(u)^{\frac{3}{2}} \frac{du}{3}$

= $\displaystyle \tfrac{2}{27}\int(u)^{\frac{3}{2}} du$

= $\displaystyle \tfrac{2}{27} * \frac{2}{5}(u)^{\frac{5}{2}} + c$

= $\displaystyle \tfrac{4}{135} * (2 + 3t)^{\frac{5}{2}} + c$

do not skip steps .... You are not ready for it.

Guy,

Make life easier for yourself by using substitution.

\(\displaystyle \int(\tfrac{u^2-2}{3}+2)\tfrac{2u^2}{3} du \\
=\tfrac{2}{9} \int (u^2+4)u^2 du \\
=\tfrac{2u^5}{45} + \tfrac{8u^3}{27} + c\)

Corollary:

\(\displaystyle u = \sqrt{2 + 3t} \\
du = \tfrac{3}{2}(2 + 3t)^{-\frac{1}{2}} dt \\
dt = \tfrac{2u}{3} du \\
t = \tfrac{u^2-2}{3}\)

Much easier than using parts...

Forgot to remind you to re-substitute, t for u.