Integration by subsitution (give constant value), why?

[invictor]

$\int(x^{2}-5)^{2}x dx$

By substiution:
1. $u = x^{2}-5$

2. $du = 2x dx$

3. $\frac{du}{2x}= dx$

4. $\int u^{2}x \frac{du}{2x}$

5. $\int u^{2} \frac{1}{2} du$

6. $\frac{1}{3} u^{3} \frac{1}{2}$

7. $\frac{1}{6} u^{3}$

8. $\frac{1}{6} (x^{2}-5)^{3}$

9. $\frac{1}{6} [x^{6} - 15x^{4} + 75x^{2} +125]$By normal integration factorize the from beginning

from: $\int(x^{2}-5)^{2}x dx$to $\int [x^{4} - 10x^{2} + 25 ] x dx$

then: $\int x^{5} - 10x^{3} + 25x dx$

and finally : $\frac{1}{6} [x^{6} - 15x^{4} + 75x^{2}] + C$Probably this is a easy one, i been looking on internet, but had hard time to find the right keywords for an explanation...

Question is: the one give me some kind of constant and the other i just add one.. Which one is correct? I mean both gives same result except of one provide a "real" constant value.

 

[Eynstone]

A constant should be added in the 6th step of method one.The answers tally.

 

[gb7nash]

You blundered on step 6. When integrating, you must always add an arbitrary constant after integrating. The cool thing though, is that any constant you get at the end is correct (unless more information is given). There's an infinite number of answers you can get. As long as your constant is a real number, you're fine. That's why we just add a constant C to be general.

 

[invictor]

Aha yes my fault but this doesn't explain why i get 125 value from the first method

since unfactorise $(x^{2} - 5)^{3} = x^{6} - 15x^{4} + 75x^{2} - 125$

 

[invictor]

So why does both methods differs -125/6 ?

 

[invictor]

Sorry I will correct this into new post

By substiution:
1. $u = x^{2}-5$

2. $du = 2x dx$

3. $\frac{du}{2x}= dx$

4. $\int u^{2}x \frac{du}{2x}$

5. $\int u^{2} \frac{1}{2} du$

6. $\frac{1}{3} u^{3} \frac{1}{2} + C$

7. $\frac{1}{6} u^{3} +C$

8. $\frac{1}{6} (x^{2}-5)^{3} +C$

9. $\frac{1}{6} [x^{6} - 15x^{4} + 75x^{2} +125] +C$

So comparing substitution method and normal unfactor in beginning give me different answer

$\frac{1}{6} [x^{6} - 15x^{4} + 75x^{2} +125] +C \neq \frac{1}{6} [x^{6} - 15x^{4} + 75x^{2}] + C$

Where a value of 126/6 appears (a constant), so the substitution method gives me "2" constants? but 126/6 + C = C ?

Or did i make any mistake somewhere? I mean I know C can be anything sure, but shoulnt C be same for both methods?