Integral of 5x^2 + √x - 4/x^2: Step-by-Step Solution Guide

[LDC1972]

Homework Statement

Find the integral of

Homework Equations

∫(5x^2 + √x - 4/x^2 dx

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

 

[CAF123]

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

 

[LDC1972]

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

Thanks for fast reply.

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

 

[CAF123]

So do I add 1 to exponent

Yes, but 1/2 + 1 ≠ 2/3.

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

 

[LDC1972]

Thanks for fast reply.

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

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

 

[jhosamelly]

Homework Statement

Find the integral of

Homework Equations

∫(5x^2 + √x - 4/x^2 dx

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

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

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

 

[CAF123]

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

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

 

[LDC1972]

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

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

 

[jhosamelly]

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

All you need to do now is simplify

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