What's Wrong with My Integration by Parts?

  • Context: Undergrad 
  • Thread starter Thread starter QuickQThrowaway
  • Start date Start date
  • Tags Tags
    Integral Weird
Click For Summary
SUMMARY

The integral ∫x*(x-2)-2dx can be solved using integration by parts, yielding the result -x*(x-2)-1 + ln(x-2) + C. However, online calculators provide a different form: -2*(x-2)-1 + ln(x-2) + C. Both results are equivalent, differing only by a constant of integration. The confusion arises from the interpretation of the constant, which led to a misunderstanding in the context of a problem requiring verification of C=2.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of constants of integration in calculus.
  • Basic skills in manipulating algebraic expressions.
NEXT STEPS
  • Study the method of integration by parts in detail.
  • Explore the properties of logarithmic functions in calculus.
  • Learn about the implications of constants of integration in indefinite integrals.
  • Practice solving integrals that involve algebraic manipulation and verification of results.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking to clarify common misconceptions in integral calculus.

QuickQThrowaway
Messages
2
Reaction score
1
Heya...

I just stumbled upon this integral while doing ODEs...

∫x*(x-2)-2dx

I used integration by parts

u=x
du/dx=1
dv/dx=(x-2)-2
v=-(x-2)-1

Thus

∫x*(x-2)-2dx=-x(x-2)-1-∫1*-(x-2)-1dx=-x*(x-2)-1+ln(x-2)+C

Now... I don't see anything wrong with this...

But every single online integration calculator as well as the ODE solution solves this integral by doing 1/(x-2)+2/(x-2)2 and they all get

-2*(x-2)-1+ln(x-2)+C

as a result...

What am I doing wrong in my integration by parts? What am I missing? Oo
 
  • Like
Likes   Reactions: eifphysics
Physics news on Phys.org
There is nothing wrong. The two results are equivalent.

[itex]\frac{x}{x-2}=\frac{x-2+2}{x-2}=1+\frac{2}{x-2}[/itex].

The difference is only in the constant of integration.
 
  • Like
Likes   Reactions: QuickQThrowaway and mrnike992
mathman said:
There is nothing wrong. The two results are equivalent.

[itex]\frac{x}{x-2}=\frac{x-2+2}{x-2}=1+\frac{2}{x-2}[/itex].

The difference is only in the constant of integration.

Thank you!
The maths department screwed up then because the question asked to verify that C=2 but with the above integration it equals 3... So they omitted one result... >.<
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K