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 $5x^3 / 1 + 16x^4$

what i have so far is...

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

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 $5x^3 / 1 + 16x^4$

what i have so far is...

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

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.

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

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

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 $5x^3 / 1 + 16x^4$

what i have so far is...

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

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 $\int \frac{5x^3}1 + 16x^4$ So maybe next time write it as $5x^3 /(1 + 16x^4)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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