Solving the Integral: \int \dfrac{x-1}{(x+1)^3} e^x dx

[utkarshakash]

Homework Statement

\int \dfrac{x-1}{(x+1)^3} e^x dx

The Attempt at a Solution

The most I can do is split the terms and integrate them individually but I am facing problems integrating the individual terms.

 

[Saitama]

Homework Statement

\int \dfrac{x-1}{(x+1)^3} e^x dx

The Attempt at a Solution

The most I can do is split the terms and integrate them individually but I am facing problems integrating the individual terms.

Do you know about the following:
\int e^x(f(x)+f'(x))=e^xf(x)+C?
The problem becomes really easy with that.

 

[utkarshakash]

Do you know about the following:
\int e^x(f(x)+f'(x))=e^xf(x)+C?
The problem becomes really easy with that.

I already know this but I'm finding it difficult to reduce the question to this form. Can you please give me some hints.

 

[utkarshakash]

Do you know about the following:
\int e^x(f(x)+f'(x))=e^xf(x)+C?
The problem becomes really easy with that.

I already know this but I'm finding it difficult to reduce the question to this form. Can you please give me some hints?

 

[Curious3141]

I already know this but I'm finding it difficult to reduce the question to this form. Can you please give me some hints?

Pranav's hint is a good one. But to apply it, you need to see that $x-1 = (x+1) - 2$. Split that rational expression into two, simplify, and see where you go from there.

 

[HallsofIvy]

Homework Statement

\int \dfrac{x-1}{(x+1)^3} e^x dx

The Attempt at a Solution

The most I can do is split the terms and integrate them individually but I am facing problems integrating the individual terms.

Thinking you can "split the terms and integrate them individually" indicates that you need to review integration all together!

 

[utkarshakash]

Thinking you can "split the terms and integrate them individually" indicates that you need to review integration all together!

This means according to you this step is incorrect.

\displaystyle \int \dfrac{(x+1)-2}{(x+1)^3} e^x dx \\<br /> \displaystyle \int (x+1)^{-2} e^x dx - \int \dfrac{2}{(x+1)^3} e^x dx

 

[CAF123]

This means according to you this step is incorrect.

\displaystyle \int \dfrac{(x+1)-2}{(x+1)^3} e^x dx \\<br /> \displaystyle \int (x+1)^{-2} e^x dx - \int \dfrac{2}{(x+1)^3} e^x dx

That step is correct. I think what HallsofIvy thought you meant by your statement was that you were going to integrate each 'component' of your integrand separately, ie integrate (x-1), 1/(x+1)³ and e^x separately to obtain something completely nonsensical.

 

[Curious3141]

This means according to you this step is incorrect.

\displaystyle \int \dfrac{(x+1)-2}{(x+1)^3} e^x dx \\<br /> \displaystyle \int (x+1)^{-2} e^x dx - \int \dfrac{2}{(x+1)^3} e^x dx

Correct, but you don't actually have to split it up this way. Just leave the integrand as e^x times that expression, then use Pranav's hint.

 

[utkarshakash]

Correct, but you don't actually have to split it up this way. Just leave the integrand as e^x times that expression, then use Pranav's hint.

I already know that. But HallsOfIvy thought splitting the terms like this is completely wrong.

 

[Ray Vickson]

I already know that. But HallsOfIvy thought splitting the terms like this is completely wrong.

No, he did not think or say that! He just said you need to review integration. Sometimes splitting up an integral is not helpful, even though it may be correct.

 

[sankalpmittal]

I already know that. But HallsOfIvy thought splitting the terms like this is completely wrong.

Plain and simple. Write x-1=(x+1)-2 and hence break the denominator. Now you get two separate integrands. Think. Integrating anyone of the two by parts should cancel other automatically and you might be able to get the correct integral.

And do not blame others.

 

[utkarshakash]

Plain and simple. Write x-1=(x+1)-2 and hence break the denominator. Now you get two separate integrands. Think. Integrating anyone of the two by parts should cancel other automatically and you might be able to get the correct integral.

And do not blame others.

I am not blaming others. I have already arrived at the answer.