Integrate $\int\frac{x^3}{\sqrt{x^2 + 1}} dx$ from 0 to 1

[Zerkor]

Homework Statement

find the definite integral \int\frac{x^3}{\sqrt{x^2 + 1}} dx from 0 to 1

Homework Equations

The Attempt at a Solution

This problem is in the integration by parts section .. I chose u = x^3 , and dv=\frac{1}{\sqrt{x^2 + 1}} so v = \frac{x^4}{4} and du = -(x^2 + 1)^(\frac{-3}{2}) , so the integral is equal to \frac{x^4}{4} . \frac{1}{√x^2 + 1} - \int\frac{x^4}{4} . -(x^2 + 1)^(\frac{-3}{2}) .. and my problem is the integral on the right hand side of the equation; I don't know how to integrate it and I don't know whether if I've chosen the parts correctly or there is a better way of choosing the parts

 

[hilbert2]

How about first doing a change of variables s=x². Now ds=2xdx and the integrand simplifies considerably.

 

[STEMucator]

If the change of variable doesn't float your boat, you could use a trig substitution of $x = atanθ$.

 

[LCKurtz]

It isn't the easiest way to work the problem, but if you really want to use integration by parts, try$$
u = x^2,\,dv=\frac x {\sqrt{x^2+1}}$$

 

[SteamKing]

In the OP, if u = x^3, then du = 3x^2 dx, not that weird expression. This is basic differentiation.

And if dv = dx/SQRT(x^2+1), then v certainly is not equal to (1/4)*x^4.

You should review the formulas for derivatives and simple integrals.

 

[HallsofIvy]

Homework Statement

find the definite integral \int\frac{x^3}{\sqrt{x^2 + 1}} dx from 0 to 1

Homework Equations

The Attempt at a Solution

This problem is in the integration by parts section .. I chose u = x^3 , and dv=\frac{1}{\sqrt{x^2 + 1}} so v = \frac{x^4}{4} and du = -(x^2 + 1)^(\frac{-3}{2}

You have "u" and "v" confused. With u= x^3, du= 3x^2dx- you differentiated v. With dv= \frac{1}{\sqrt{x^2+ 1}}, v= arctan(x)- you integrated u.

, so the integral is equal to \frac{x^4}{4} . \frac{1}{√x^2 + 1} - \int\frac{x^4}{4} . -(x^2 + 1)^(\frac{-3}{2}) .. and my problem is the integral on the right hand side of the equation; I don't know how to integrate it and I don't know whether if I've chosen the parts correctly or there is a better way of choosing the parts

 

[pashav46]

A one solve the definite integral

∫x²/√x²+1 dx={z=x²+1,dz=2*x*dx}=
∫(z-1)/(2√z) dz=1/3*z^3/2-√z={x=0⇔z=1,x=1⇔z=2}=(2-√2)/3

 

[LCKurtz]

All these responses and suggestions, and Zerkor never returned to his thread.

 

[Zerkor]

It isn't the easiest way to work the problem, but if you really want to use integration by parts, try$$
u = x^2,\,dv=\frac x {\sqrt{x^2+1}}$$

I tried this one and it didn't work .. In the sections I've studied in the previous two days I learned trigonometric substitution and miscellaneous substitution; I think I have to try working this problem with those techniques

 

[Zerkor]

Thanks a lot guys for your help and I'm sorry for replying after a long period but I didn't access the forum through those two days :)

 

[SteamKing]

Without showing your work, we can't say why integration by parts didn't work for you, except you must have made a mistake. IBP does work for this integral, but your earlier postings show that you do not have a grasp on how to apply it correctly.

 

[Zerkor]

Without showing your work, we can't say why integration by parts didn't work for you, except you must have made a mistake. IBP does work for this integral, but your earlier postings show that you do not have a grasp on how to apply it correctly.

May be you are right. But I've just worked it using trigonometric substitution using x = tanθ and it worked correctly .. Maybe there exist a way to find it using IBP but it's a hard way and there is no use for it if there is an easier way :)

 

[LCKurtz]

May be you are right. But I've just worked it using trigonometric substitution using x = tanθ and it worked correctly .. Maybe there exist a way to find it using IBP but it's a hard way and there is no use for it if there is an easier way :)

But since you are learning calculus, don't you think it might be of some value to figure out why it's not working for you?