invictor
- 6
- 0
\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 dxto \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}] + CProbably 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.
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 dxto \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}] + CProbably 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.
Last edited: