Substitution Integral

**Em Yeu Anh** · January 22nd 2010, 06:17 PM

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

I am stumped on what to substitute here.

**VonNemo19** · January 22nd 2010, 06:52 PM

Originally Posted by Em Yeu Anh

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

I am stumped on what to substitute here.

EDIT: uh sorry.

**General** · January 22nd 2010, 06:53 PM

Originally Posted by Em Yeu Anh

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

I am stumped on what to substitute here.

Try $x=u^{12}$ .

**simplependulum** · January 24th 2010, 02:10 AM

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

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

the integral goes to

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

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

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

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

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

and so on

**Em Yeu Anh** · January 25th 2010, 05:33 PM

Originally Posted by simplependulum

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

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

the integral goes to

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

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

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

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

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

and so on

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

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

**simplependulum** · January 25th 2010, 11:06 PM

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

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

**Em Yeu Anh** · January 26th 2010, 04:04 PM

Originally Posted by simplependulum

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

Is that not the same as what I just put?

**General** · January 26th 2010, 04:07 PM

Originally Posted by Em Yeu Anh

Is that not the same as what I just put?

No. As an counter-example, put $u=2$ in both of them, and you will see that they are not equal to each other.
In general: $u^a \neq u^{-a}$ .

**Em Yeu Anh** · January 26th 2010, 08:04 PM

My eyes completely missed that negative sign, sorry.

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