Integrate by Parts: x^5 * sqrt(x^3 + 5)

[whatlifeforme]

Homework Statement

Integrate by parts.

Homework Equations

(integral) (x^5 * sqrt(x^3 + 5) dx)

The Attempt at a Solution

i've tried using simple substitution, not by parts.

integral (x^3 * x^2 * sqrt(x^3 + 5) dx
u=x^3 + 5
du=3x^2

1/3(integral) (u-5) * u^1/2 du
1/3(u^3/2 - 5u^1/2)
--------------------------------------------

2 (x^3 + 5)^5/2
--
15

(subtract)

10 (x^3 + 5)^3/2
--
3

 

[SammyS]

Homework Statement

Integrate by parts.

Homework Equations

(integral) (x^5 * sqrt(x^3 + 5) dx)

The Attempt at a Solution

i've tried using simple substitution, not by parts.

integral (x^3 * x^2 * sqrt(x^3 + 5) dx
u=x^3 + 5
du=3x^2

1/3(integral) (u-5) * u^1/2 du

How did you get the (u-5) ?

1/3(u^3/2 - 5u^1/2)
--------------------------------------------

2 (x^3 + 5)^5/2
--
15

(subtract)

10 (x^3 + 5)^3/2
--
3

 

[Mark44]

Homework Statement

Integrate by parts.

Homework Equations

(integral) (x^5 * sqrt(x^3 + 5) dx)

The Attempt at a Solution

i've tried using simple substitution, not by parts.

integral (x^3 * x^2 * sqrt(x^3 + 5) dx
u=x^3 + 5
du=3x^2

1/3(integral) (u-5) * u^1/2 du
1/3(u^3/2 - 5u^1/2)
--------------------------------------------

2 (x^3 + 5)^5/2
--
15

(subtract)

10 (x^3 + 5)^3/2
--
3

What's your question?

You can always check your answer to an integration problem by differentiating your answer. If this results in the integrand, then your answer is correct.

 

[whatlifeforme]

How did you get the (u-5) ?

u=x^3 + 5 --> u-5=x^3

to substitute for the x^3 in the equation i rearrange the above by subtracting 5 from both sides.

 

[STEMucator]

Homework Statement

Integrate by parts.

Homework Equations

(integral) (x^5 * sqrt(x^3 + 5) dx)

The Attempt at a Solution

i've tried using simple substitution, not by parts.

integral (x^3 * x^2 * sqrt(x^3 + 5) dx
u=x^3 + 5
du=3x^2

1/3(integral) (u-5) * u^1/2 du
1/3(u^3/2 - 5u^1/2)
--------------------------------------------

2 (x^3 + 5)^5/2
--
15

(subtract)

10 (x^3 + 5)^3/2
--
3

u = x^3 + 5, so (1/3)du = x^2dx.

Notice you can write x^3 = u-5?

I'm getting : $\frac{2}{15}(x^3+5)^{5/2} - \frac{10}{9}(x^3+5)^{3/2} + c$

 

[whatlifeforme]

u = x^3 + 5, so (1/3)du = x^2dx.

Notice you can write x^3 = u-5?

I'm getting : $\frac{2}{15}(x^3+5)^{5/2} - \frac{10}{9}(x^3+5)^{3/2} + c$

they have as the answer: (2/9)x^3 * (x^3 + 5)^(3/2) - (4/45)(x^3 + 5)^(5/2) + C.

 

[Dick]

they have as the answer: (2/9)x^3 * (x^3 + 5)^(3/2) - (4/45)(x^3 + 5)^(5/2) + C.

Their answer is right and Zondrina's answer is also right. Your's isn't right. There's more than one way of writing an expression. Differentiate your answer to see if it's correct.

 

[whatlifeforme]

how are those two expressions the same?

 

[Dick]

how are those two expressions the same?

Factor (x^3+5)^(3/2) out of both expressions and rearrange what's left.

 

[whatlifeforme]

u = x^3 + 5, so (1/3)du = x^2dx.

Notice you can write x^3 = u-5?

I'm getting : $\frac{2}{15}(x^3+5)^{5/2} - \frac{10}{9}(x^3+5)^{3/2} + c$

can you please show how?

 

[whatlifeforme]

bump.

 

[Dick]

bump.

Again, differentiate your answer and see what you get. Then realize you didn't multiply your second term by 1/3. The coefficient should be -10/9, not -10/3. Your result is almost correct, it's just a simple mistake.

 

[whatlifeforme]

how would i solve this using integration by parts as opposed to u-sub like the problem states.

 

[iRaid]

I just did it and you can make it work by doing u=x^3 and then you end up with
$$\int u\sqrt{u+5}du$$
So then you do an integration by parts and it works out.

 

[SammyS]

how would i solve this using integration by parts as opposed to u-sub like the problem states.

There are two reasonable ways to assign pair of functions, u(x) and v(x) so that $\displaystyle \ \ u(x)v'(x)=x^5\sqrt{x^3+5}\ .$

One such assignment is

$\displaystyle \ \ u(x)=\sqrt{x^3+5}\,,\$
and
$\displaystyle \ \ v'(x)=x^5\ .$

The resulting integral has an integrand of $\displaystyle \ \ v(x)u'(x)\,, \$ which has x raised to a higher power than the original and a radical in the denominator.

This is probably not a good choice.​

The other assignment can be most easily obtained by first assigning v(x), then finding v'(x), with u(x) being what's "left over".

Let $\displaystyle \ \ v(x)=(x^3+5)^{3/2}\,,\$ so that v'(x) includes $\displaystyle \ \ \sqrt{x^3+5}\ .$

This gives $\displaystyle \ \ v'(x)=\frac{9}{2}x^2\sqrt{x^3+5}\ .\$

That leaves you with $\displaystyle \ \ u(x)=\frac{2}{9}x^3\ .$​

See where this second assignment for u(x) and v(x) leads you.