1. ## integration by substitution...quotient

i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

integrate $\displaystyle 5x^3 / 1 + 16x^4$

what i have so far is...

$\displaystyle Let u = 1 + 16x^4$
$\displaystyle du/dx = 64x^3$
$\displaystyle du = 64x^3 dx$

now what do i do??

i know that it has to be 5/u and then something.....

2. Originally Posted by b00yeah05
i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

integrate $\displaystyle 5x^3 / 1 + 16x^4$

what i have so far is...

$\displaystyle Let u = 1 + 16x^4$
$\displaystyle du/dx = 64x^3$
$\displaystyle du = 64x^3 dx$

now what do i do??

i know that it has to be 5/u and then something.....
The good news is that you're almost on exactly the right track.

$\displaystyle du = 64x^3 dx \Rightarrow dx = \frac{du}{64x^3}$

So the integral becomes $\displaystyle \int \frac{5x^3}{u} \, \frac{du}{64x^3} = \int \frac{5}{64} \, \frac{1}{u} \, du = \frac{5}{64} \int \frac{1}{u} \, du = ......$

3. yes..thanks mr fantastic i worked that one out eventually..but just a general question on integration by substitution now.. do you always have to times by 'du' in the end? for all questions??

4. Originally Posted by b00yeah05
yes..thanks mr fantastic i worked that one out eventually..but just a general question on integration by substitution now.. do you always have to times by 'du' in the end? for all questions??
You have to substitute for x AND dx. The expression for dx will always contain a du. So yes, I guess that means you will always have to "times by a du in the end" ...

5. Originally Posted by b00yeah05
i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

integrate $\displaystyle 5x^3 / 1 + 16x^4$

what i have so far is...

$\displaystyle Let u = 1 + 16x^4$
$\displaystyle du/dx = 64x^3$
$\displaystyle du = 64x^3 dx$

now what do i do??

i know that it has to be 5/u and then something.....
You should use parentheses, your question looks like $\displaystyle \int \frac{5x^3}1 + 16x^4$ So maybe next time write it as $\displaystyle 5x^3 /(1 + 16x^4)$

Anyway,
$\displaystyle \int \frac{5x^3}{1 + 16x^4}dx$

5 is a coefficient, so you can pull it to the outside
$\displaystyle =5\int \frac{x^3}{1 + 16x^4}dx$

Now partition your integral so that it is easier to see what you are doing
$\displaystyle \begin{array}{rcc} =5\int &\frac{1}{1 + 16x^4}&x^3~dx\\ \end{array}$

$\displaystyle \begin{array}{|c|} \hline SUBSTITUTION\\\hline\\ u=1+16x^4\\\\ du=64x^3dx\\\\ \frac 1{64}du=x^3dx\\\\\hline\end{array}$

Determine what will substitute with what
$\displaystyle \begin{array}{rcc} &&\frac 1{64}~du\\ =5\int &\underbrace{\frac{1}{1 + 16x^4}}&\overbrace{x^3~dx}\\ &u \end{array}$

So make the substitution
$\displaystyle =5\int \frac{1}{u}*\frac{1}{64}~du$

1/64 is a constant, so it is part of the coefficient, so move it to the front to get it out of your way.
$\displaystyle =5*\frac{1}{64}\int \frac{1}{u}~du$

$\displaystyle =\frac{5}{64}\int \frac{1}{u}~du$

Now, you should have memorized that $\displaystyle \int \frac 1x ~dx = ln|x|+C$ so apply this:

$\displaystyle =\frac{5}{64}~ln|u|+C$

And now reverse substitute from u back into x:
$\displaystyle =\frac{5}{64}~ln|1+16x^4|+C$

edit: beat me to it >.< Darn arrays

6. Originally Posted by b00yeah05
i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

integrate $\displaystyle 5x^3 / 1 + 16x^4$
Based on the idea where $\displaystyle \int {\frac{{f'(x)}} {{f(x)}}\,dx} = \ln \left| {f(x)} \right| + k,\,f(x) \ne 0,$ you just turn the integral into that form:

$\displaystyle \int {\frac{{5x^3 }} {{1 + 16x^4 }}\,dx} = \frac{5} {{64}}\int {\frac{{\left( {1 + 16x^4 } \right)'}} {{1 + 16x^4 }}\,dx} = \frac{5} {{64}}\ln \left( {1 + 16x^4 } \right) + k.$

(Absolute value is omitted here since $\displaystyle 1+16x^4$ it's always positive.)

P.S.: see my signature for better LaTeX typesetting.