Thread: Integrate

1. Integrate

$\displaystyle \int \frac{x}{25+16x^2} dx$

$\displaystyle u=25+16x^2$

$\displaystyle du=32xdx$

$\displaystyle \frac{1}{32}du=x dx$

$\displaystyle \frac{1}{32}\int \frac{1}{u} du \rightarrow \frac{1}{32}[1] \rightarrow \frac{1}{32} + C$

Where did I go wrong or miss something?

2. from (1/32)S(1/u).du * = (1/32)ln(u)

then sub u in?

= (1/32)ln(25+16(x^2))

= [ln(25+16(x^2))]/32

Is it this?

*the S is the integral sign

3. Originally Posted by likearollingstone
from (1/32)S(1/u).du * = (1/32)ln(u)

then sub u in?

= (1/32)ln(25+16(x^2))

= [ln(25+16(x^2))]/32

Is it this?

*the S is the integral sign
I think i see my mistake $\displaystyle \int \frac{1}{x}dx \rightarrow ln|x| + C$ right?

4. You have to substitute back for u. So it would be

(ln(25+16x^2))/32

5. Originally Posted by VitaX
I think i see my mistake $\displaystyle \int \frac{1}{x}dx \rightarrow ln|x| + C$ right?
Yep thats right

6. Originally Posted by emeryj
You have to substitute back for u. So it would be

(ln(25+16x^2))/32
Yah I know you have to sub back but why are you writing parenthesis isnt it absolute value?

7. Originally Posted by VitaX
Yah I know you have to sub back but why are you writing parenthesis isnt it absolute value?
25+16(x^2) is always greater than zero so you dont really need the absolute sign but i give you permision to put it in anyway.