# Thread: Substitution Integral

1. ## Substitution Integral

$\displaystyle \int\frac{dx}{(1+x^{\frac{1}{4}})x^{\frac{1}{3}}}$

I am stumped on what to substitute here.

2. Originally Posted by Em Yeu Anh
$\displaystyle \int\frac{dx}{(1+x^{\frac{1}{4}})x^{\frac{1}{3}}}$

I am stumped on what to substitute here.
EDIT: uh sorry.

3. Originally Posted by Em Yeu Anh
$\displaystyle \int\frac{dx}{(1+x^{\frac{1}{4}})x^{\frac{1}{3}}}$

I am stumped on what to substitute here.
Try $\displaystyle x=u^{12}$.

4. Try to sub. $\displaystyle x = u^{-12}$

$\displaystyle dx = -12u^{-13} du$

the integral goes to

$\displaystyle \int \frac{-12u^{-13} du}{ ( 1+ u^{-12/4}) u^{-12/3} }$

$\displaystyle = \int \frac{-12u^{-13} du}{ ( 1+ u^{-3}) u^{-4} }$

$\displaystyle = -12 \int \frac{du}{u^6(1 + u^3)}$

$\displaystyle = - 12 \int \left[ \frac{1}{1+u^3} - \frac{u^3- 1}{u^6} \right] ~du$

$\displaystyle = -12 \int \left[ \frac{1}{1 + u^3 } - \frac{1}{u^3} + \frac{1}{u^6} \right] ~du$

and so on

5. Originally Posted by simplependulum
Try to sub. $\displaystyle x = u^{-12}$

$\displaystyle dx = -12u^{-13} du$

the integral goes to

$\displaystyle \int \frac{-12u^{-13} du}{ ( 1+ u^{-12/4}) u^{-12/3} }$

$\displaystyle = \int \frac{-12u^{-13} du}{ ( 1+ u^{-3}) u^{-4} }$

$\displaystyle = -12 \int \frac{du}{u^6(1 + u^3)}$

$\displaystyle = - 12 \int \left[ \frac{1}{1+u^3} - \frac{u^3- 1}{u^6} \right] ~du$

$\displaystyle = -12 \int \left[ \frac{1}{1 + u^3 } - \frac{1}{u^3} + \frac{1}{u^6} \right] ~du$

and so on
Thanks! that's an interesting way to solve it! would never have thought of that myself.

Only one question, $\displaystyle = -12 \int \frac{du}{u^6(1 + u^3)}$ should that instead be $\displaystyle = -12 \int \frac{du}{u^9(1 + u^3)}$ or am I wrong?

6. Originally Posted by Em Yeu Anh
Thanks! that's an interesting way to solve it! would never have thought of that myself.

Only one question, $\displaystyle = -12 \int \frac{du}{u^6(1 + u^3)}$ should that instead be $\displaystyle = -12 \int \frac{du}{u^9(1 + u^3)}$ or am I wrong?

Be careful of the denominator which is $\displaystyle u^9(1 + u^{-3})$

7. Originally Posted by simplependulum
Be careful of the denominator which is $\displaystyle u^9(1 + u^{-3})$
Is that not the same as what I just put?

8. Originally Posted by Em Yeu Anh
Is that not the same as what I just put?
No. As an counter-example, put $\displaystyle u=2$ in both of them, and you will see that they are not equal to each other.
In general: $\displaystyle u^a \neq u^{-a}$.

9. My eyes completely missed that negative sign, sorry.