Find integral using substition: int ( 10x^2 (3 + 4x^2)^(1/4) ) dx

[Themrc]

I can't figure this problem out. It doesn't appear to allow substitution because when I set u=3+4x^3, du=12x^2dx.

That isn't 10x^2dx. Can someone explain how I can make this problem work using substitution.

The problem is #28 in the photo. Thanks!

 

[Deleted member 4993]

I can't figure this problem out. It doesn't appear to allow substitution because when I set u=3+4x^3, du=12x^2dx.

That isn't 10x^2dx. Can someone explain how I can make this problem work using substitution.

The problem is #28 in the photo. Thanks!

Sure. But

du = 12x^2 dx

10 x^2 dx = 10/12 * du

as simple as that.......

 

[stapel]

I can't figure this problem out. It doesn't appear to allow substitution because when I set u = 3 + 4x^3, [then I get] du = 12x^2 dx.

[But t]hat [expression] isn't [the same as what I think I'm needing, which is] 10x^2dx. Can someone explain how I can make this problem work using substitution[?]

The problem is #28 in the photo.

In future, please type out the exercise in question:

. . . . .28) int ( 10x^2 (3 + 4x^2)^(1/4) ) dx

...or at least please make sure that the image is right-side-up, so that it can be read easily. Thank you!

 

[Prove It]

I can't figure this problem out. It doesn't appear to allow substitution because when I set u=3+4x^3, du=12x^2dx.

That isn't 10x^2dx. Can someone explain how I can make this problem work using substitution.

The problem is #28 in the photo. Thanks!

$\displaystyle \begin{align*} \int{ 10\,x^2 \left( 3 + 4\,x^3 \right) ^{\frac{1}{4}} \,\mathrm{d}x } &= \frac{5}{6} \int{ 12\,x^2 \left( 3 + 4\,x^3 \right) ^{\frac{1}{4}}\,\mathrm{d}x } \\ &= \frac{5}{6} \int{ u^{\frac{1}{4}}\,\mathrm{d}u } \, \, \textrm{ when we substitute} u = 3 + 4\,x^3 \implies \mathrm{d}u = 12\,x^2 \\ &= \frac{5}{6} \left( \frac{4}{5} \, u^{\frac{5}{4}} \right) + C \\ &= \frac{2}{3}\,u^{\frac{5}{4}} + C \\ &= \frac{2}{3} \left( 3 + 4\,x^3 \right) ^{\frac{5}{4}} + C \end{align*}$

 

[stapel]

$\displaystyle \displaystyle \int\, 10\,x^2 \left(3\, + \,4\,x^3 \right) ^{\frac{1}{4}}\, dx\, =\,$$\displaystyle \dfrac{5}{6}\,$$\displaystyle \displaystyle \int\, 12\, x^2\, \left(3\, +\, 4\, x^3\right)^{\frac{1}{4}}\, dx$

In future, before merely posting a solution, please read the poster's questions. The bits you skipped between the left-hand side of the above equality and the right-hand side are exactly what were causing the student's difficulty. :shock:

 

[Prove It]

In future, before merely posting a solution, please read the poster's questions. The bits you skipped between the left-hand side of the above equality and the right-hand side are exactly what were causing the student's difficulty. :shock:

I showed EXACTLY how to get it into a form to use a substitution. The point I was making was that substitution IS NOT ENOUGH!