Indefinite integral

**bgwarhawk** · December 11th 2008, 05:39 PM

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

Thanks

**Skalkaz** · December 11th 2008, 06:01 PM

Dear bgwarhawk,

these are elementary intagrals. You must recall the formula:

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

**bgwarhawk** · December 11th 2008, 06:09 PM

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?

**Skalkaz** · December 11th 2008, 06:20 PM

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?

**bgwarhawk** · December 11th 2008, 06:26 PM

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...

**Skalkaz** · December 11th 2008, 07:00 PM

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

**Erdos32212** · December 11th 2008, 07:44 PM

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

**Prove It** · December 11th 2008, 07:48 PM

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 $\int x^a dx = \frac{1}{a + 1}x^{a+1}+C$ .

**Prove It** · December 11th 2008, 07:49 PM

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. $\int{7x^3 - 4x^2 + 9x - 10\,dx} = \frac{7}{4}x^4 - \frac{4}{3}x^3 + \frac{9}{2}x^2 - 10x + C$

**Prove It** · December 11th 2008, 07:51 PM

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?

**bgwarhawk** · December 12th 2008, 08:06 AM

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?

**Prove It** · December 13th 2008, 01:44 AM

Originally Posted by bgwarhawk

What do I do?

Use the formula $\int x^a dx = \frac{1}{a + 1}x^{a+1}+C$ .

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

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

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

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

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