Integration with power substitution

[gruba]

Homework Statement

Find the integral $\int \frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$

2. The attempt at a solution
$$\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}=\frac{2\sqrt[5]{2x-3}-1}{2\sqrt[5]{2x-3}(x-1)}=\frac{1}{x-1}-\frac{1}{2\sqrt[5]{2x-3}(x-1)}$$$$\int \frac{1}{x-1}\mathrm dx=\ln|x-1|+c$$
$$\int \frac{1}{2\sqrt[5]{2x-3}(x-1)}\mathrm dx=\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx$$

Substitution $u=2x-3,du=2dx$ gives
$$\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$$

Substitution $u^{1/5}=v,u=v^5,du=5v^4$ gives
$$\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du=5\int \frac{v^3}{v^5+1}\mathrm dv$$

$$v^5+1=(v+1)(v^4-v^3+v^2-v+1)$$

Using partial fractions:
$$\frac{v^3}{v^5+1}=\frac{A}{v+1}+\frac{Bv^3+Cv^2+Dv+E}{v^4-v^3+v^2-v+1}$$

$$\Rightarrow A=-1/3,B=1/3,C=1,D=-2/3,E=1/3$$

$$\int \frac{A}{v+1}\mathrm dv=-\frac{1}{3}\ln|v+1|+c$$

How to integrate $$\frac{Bv^3+Cv^2+Dv+E}{v^4-v^3+v^2-v+1}?$$

 

[Ssnow]

There is an other half when you substitute $2x-3=u$ because $dx=\frac{du}{2}$, but this not the principal problem ... other things seems correct, for your last integral you must factorize the polynomial $v^4-v^{3}+v^{2}-v+1$ that has four solution in $\mathbb{C}$ and proceed with the partial fractions in the complex case ...

 

[gruba]

There is an other half when you substitute $2x-3=u$ because $dx=\frac{du}{2}$, but this not the principal problem ... other things seems correct, for your last integral you must factorize the polynomial $v^4-v^{3}+v^{2}-v+1$ that has four solution in $\mathbb{C}$ and proceed with the partial fractions in the complex case ...

I found the easier method to integrate $\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx$.
$$\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int (x-1)^{-1}(-3+2x)^{-1/5}\mathrm dx$$
which is the integral of differential binomial.

After substitution $u=x-1,du=dt$,
$$\int u^{-1}(-1+2u)^{-1/5}\mathrm du$$

which can be solved by substitution $-1+2u=v^5,du=\frac{5v^4}{2}dv$
$$\Rightarrow \int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=10\int \frac{v^4}{v^5+1}\mathrm dv$$

 

[Samy_A]

I found the easier method to integrate $\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx$.
$$\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int (x-1)^{-1}(-3+2x)^{-1/5}\mathrm dx$$
which is the integral of differential binomial.

After substitution $u=x-1,du=dt$,
$$\int u^{-1}(-1+2u)^{-1/5}\mathrm du$$

which can be solved by substitution $-1+2u=v^5,du=\frac{5v^4}{2}dv$
$$\Rightarrow \int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=10\int \frac{v^4}{v^5+1}\mathrm dv$$

Didn't you loose the $v$ from $(-1+2u)^{-1/5}$ in the denominator?

 

[gruba]

Didn't you loose the $v$ from $(-1+2u)^{-1/5}$ in the denominator?

You are right, it should be $10\int \frac{v^4}{(v^5+1)^2}\mathrm dv$.

 

[Samy_A]

You are right, it should be $10\int \frac{v^4}{(v^5+1)^2}\mathrm dv$.

No. What is $(-1+2u)^{-1/5}$ in terms of $v$, if $-1+2u=v^5$?

 

[gruba]

No. What is $(-1+2u)^{-1/5}$ in terms of $v$, if $-1+2u=v^5$?

$(-1+2u)^{-1/5}=v^{-1}\Rightarrow 5\int \frac{v^3}{v^5+1}\mathrm dv$

 

[Samy_A]

$(-1+2u)^{-1/5}=v^{-1}\Rightarrow 5\int \frac{v^3}{v^5+1}\mathrm dv$

You again lost a factor 2, as @Ssnow noticed.
I think the correct integral is $10 \int \frac{v^3}{v^5+1}\mathrm dv$

 

[gruba]

Thanks for the help.