1. ## Indefinite integral

1) ∫(7x^(3) - 4x^(2) +9x -10)dx
2) ∫ [(6 √(x)+2)/ √(x)]
3) ∫6x^(7) +4 / (x)

Thanks

2. Dear bgwarhawk,

these are elementary intagrals. You must recall the formula:

$\int x^a dx = x^{a+1}/a+C$ (if a <> -1)

3. Originally Posted by Skalkaz
Dear bgwarhawk,

these are elementary intagrals. You must recall the formula:

$\int x^a dx = x^{a+1}/a+C$ (if a <> -1)
What in the world is that?

4. This is a formula from second page in a book about integral. What is your problem? Where did you get stuck? Or do you think this forum is a automatic solution generator of homework?

5. Originally Posted by Skalkaz
This is a formula from second page in a book about integral. What is your problem? Where did you get stuck? Or do you think this forum is a automatic solution generator of homework?
For the 1) I got 7/2x^-4/3x^3+9/2x^2-10x but thats not right...

6. the 1st term is incorrect. And don't forget the C at the end.

7. Originally Posted by bgwarhawk
1) ∫(7x^(3) - 4x^(2) +9x -10)dx
2) ∫ [(6 √(x)+2)/ √(x)]
3) ∫6x^(7) +4 / (x)

Thanks
1) ∫(7x^(3) - 4x^(2) +9x -10)dx= $\frac{7x^4}{4}-\frac{4x^3}{3}+\frac{9x^2}{2}-10+C$

2) divide $\sqrt{x}$ into both terms

3)power rule and natural log

8. Originally Posted by Skalkaz
Dear bgwarhawk,

these are elementary intagrals. You must recall the formula:

$\int x^a dx = x^{a+1}/a+C$ (if a <> -1)
Actually it's $\int x^a dx = \frac{1}{a + 1}x^{a+1}+C$.

9. Originally Posted by Erdos32212
1) ∫(7x^(3) - 4x^(2) +9x -10)dx= $\frac{7x^4}{4}-\frac{4x^3}{3}+\frac{9x^2}{2}-10+C$

2) divide $\sqrt{x}$ into both terms

3)power rule and natural log
1. $\int{7x^3 - 4x^2 + 9x - 10\,dx} = \frac{7}{4}x^4 - \frac{4}{3}x^3 + \frac{9}{2}x^2 - 10x + C$

10. Originally Posted by bgwarhawk
1) ∫(7x^(3) - 4x^(2) +9x -10)dx
2) ∫ [(6 √(x)+2)/ √(x)]
3) ∫6x^(7) +4 / (x)

Thanks
2. $\int{\frac{6\sqrt{x} +2}{\sqrt{x}}\,dx} = \int{6 + \frac{2}{\sqrt{x}}\,dx} = \int{6 + 2x^{-\frac{1}{2}}\,dx}$

Can you go from there?

11. Originally Posted by Prove It
2. $\int{\frac{6\sqrt{x} +2}{\sqrt{x}}\,dx} = \int{6 + \frac{2}{\sqrt{x}}\,dx} = \int{6 + 2x^{-\frac{1}{2}}\,dx}$

Can you go from there?
What do I do?

12. Originally Posted by bgwarhawk
What do I do?
Use the formula $\int x^a dx = \frac{1}{a + 1}x^{a+1}+C$.

$\int{6 + 2x^{-\frac{1}{2}}\,dx} = \int{6x^0 + 2x^{-\frac{1}{2}}\,dx}$

$= \frac{6}{1}x^1 + \frac{2}{\frac{1}{2}}x^{\frac{1}{2}} + C$

$= 6x + 4x^{\frac{1}{2}} + C$

$= 6x + 4\sqrt{x} +C$.