What is the smart substitution for solving this integral?

[XtremePhysX]

Homework Statement

Find:

Homework Equations

\int \frac{dx}{1+x^{\frac{1}{4}}}

The Attempt at a Solution

I tried partial fractions and substitution, did work.
Tried to do it with a contour integral didn't work.
please show all working.

 

[XtremePhysX]

Do we split the integral by completing the square?

 

[micromass]

I tried partial fractions and substitution, did work.
Tried to do it with a contour integral didn't work.

Can you show us EXACTLY what you tried??

please show all working.

No, we will not. Read the forum rules. We will NOT provide you with a solution. You will have to find it yourself using our hints.

 

[XtremePhysX]

\int \frac{1}{(1+\sqrt[4]{x})^{2}-2\sqrt{x}}
then partial fractions to split the integral but it got tedious and I am not sure if I am on the right track, i need the solution asap, please help me.

 

[micromass]

\int \frac{1}{(1+\sqrt[4]{x})^{2}-2\sqrt{x}}
then partial fractions to split the integral but it got tedious and I am not sure if I am on the right track,

You can simplofy the integral to an integral of a rational functon

\int \frac{1}{1+\sqrt[4]{x}}dx

by applying a substitution. Do you see an easy substitution that you can do?

i need the solution asap, please help me.

Please don't post stuff like this. If you want help fast, you should have posted sooner. It's not nice of you to push people like this.

 

[XtremePhysX]

You can simplofy the integral to an integral of a rational functon

\int \frac{1}{1+\sqrt[4]{x}}dx

by applying a substitution. Do you see an easy substitution that you can do?



Please don't post stuff like this. If you want help fast, you should have posted sooner. It's not nice of you to push people like this.

No, not really.
What would you use as a substitution?

 

[micromass]

Take u=\sqrt[4]{x}. After that substitution, it should be easy.

 

[XtremePhysX]

I think I was able to do it with a substitution.

\int \frac{1}{1+x^{\frac{1}{4}}} dx \\ $Let u^4=x $ \therefore 4u^3 du = dx \\ I=4 \int \frac{u^3}{1+u} du\\ = 4\int u^2-u+1-\frac{1}{1+u} du $ by long division or synthetic division $ \\ = 4 \int [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]du = 4[\frac{x^{\frac{3}{4}}}{3}-\frac{x^{\frac{1}{2}}}{2}+x^{\frac{1}{4}}-\ln ( x^{\frac{1}{4}}+1)]+C

 

[XtremePhysX]

\int \frac{1}{\sqrt{e^{2x}-1}} dx

Can be done in about five easy lines with a smart substitution.

But I don't know what to use as a substitution, can you please help me?

 

[XtremePhysX]

Do I let u=e^x?

 

[dextercioby]

Well, just like above, get rid of the 'nasty' part, in this case, the exponential.

 

[micromass]

I think I was able to do it with a substitution.

\int \frac{1}{1+x^{\frac{1}{4}}} dx \\ $Let u^4=x $ \therefore 4u^3 du = dx \\ I=4 \int \frac{u^3}{1+u} du\\ = 4\int u^2-u+1-\frac{1}{1+u} du $ by long division or synthetic division $ \\ = 4 \int [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]du = 4[\frac{x^{\frac{3}{4}}}{3}-\frac{x^{\frac{1}{2}}}{2}+x^{\frac{1}{4}}-\ln ( x^{\frac{1}{4}}+1)]+C

That last line doesn't make sense. You can't say

4\int u^2-u+1-\frac{1}{1+u} du= 4 \int [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]du

You're not taking the integral anymore in the right hand side. You should write

4\int u^2-u+1-\frac{1}{1+u} du= 4 [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]

But anyway, the solution is correct!

 

[micromass]

\int \frac{1}{\sqrt{e^{2x}-1}} dx

Can be done in about five easy lines with a smart substitution.

But I don't know what to use as a substitution, can you please help me?

Please start a new thread for a new problem.