# Find the integral of:

1. Jul 18, 2013

### LDC1972

1. The problem statement, all variables and given/known data

Find the integral of

2. Relevant equations

∫(5x^2 + √x - 4/x^2 dx

3. The attempt at a solution

∫(5x^2 + √x - 4/x^2 dx

= 5 ∫x^2 dx + ∫√x dx ∫- 4/x^2
= 5 [x^3/3] + ∫√x dx ∫- 4/x^2
= 5 [x^3/3] + [x^1/2] dx ∫- 4/x^2
= 5 [x^3/3] + [x^1/2] - [-4x^-2]
= 5[x^3/3] + [x^1/2] + [4/x] + C
= 5x^3/3 + x^1/2 + 4/x + C

Does my final answer look right for this? I'm really struggling with integrals, and am hoping I'm getting to grip with it now, but if someone can verify that would be great.

Thanks so much,

Lloyd

2. Jul 18, 2013

### CAF123

$$\sqrt{x} = x^{\frac{1}{2}},$$ You must then integrate this. The rest is fine.

3. Jul 18, 2013

### LDC1972

So do I add 1 to exponent = x^2/3
Divide by same so = x^2/3 / 2/3

From there can I simplify this to just x?

Thanks

Lloyd

4. Jul 18, 2013

### CAF123

Yes, but 1/2 + 1 ≠ 2/3.

Also xa/a ≠ x, for example x2/2 = 1/2 when x = 1.

5. Jul 18, 2013

### LDC1972

Really sorry I was being dumb there...

I wrote in pen 3/2 then typed 2/3.

So to recap:

x^1/2
Add 1 to exp' = X^3/2
Divide by same = X^3/2 / 3/2

From here there must be some simplification, as the fraction is improper?

Or is that the conclusion?

Thank you,

Lloyd

6. Jul 18, 2013

### jhosamelly

$\int \sqrt{x} dx$ = $\int \ {x^{\frac{1}{2}}} dx$

= $\frac{2}{3} x^{\frac{3}{2}}$

7. Jul 18, 2013

### CAF123

Yes, you can write $$\frac{x^{\frac{3}{2}}}{3/2} = x^{\frac{3}{2}} \frac{1}{3/2} = \frac{2}{3} x^{\frac{3}{2}}$$

8. Jul 18, 2013

### LDC1972

Thank you both.

I'm still not 100% on these, but have come a long way (from knowing nothing to where I am now) in the last 7 hours.

Thankfully calculus is only a mall proportion of the study I am doing.

Thanks again,

Lloyd

9. Jul 18, 2013

### jhosamelly

All you need to do now is simplify

$\frac{5}{3}x ^{3} + \frac{2}{3} x^{\frac{3}{2}} + \frac{4}{x}$