## MarcLeclair one year ago how do you do the following integral : int (4t / (1+6t^2)) I know the answer is 1/8 ln ( 1 +6t^2) but in the process, the answer shows the numerator becoming 32t and I have no idea where the 1/8 came from.

1. Outkast3r09

$\int \frac{4t}{1+6t^2}$

2. Outkast3r09

$4\int \frac{t}{1+6t^2}dt$ $u=1+6t^2$ $du=12tdt$ $\frac{du}{12}=tdt$

3. MarcLeclair

Ah you used substitution. Thanks a lot, makes muh more sense

4. Outkast3r09

ys and it's 1/3

5. Outkast3r09

wolframalpha is a good site to use to check

6. campbell_st

ok... so it looks like a log function to start s = 1 + 6u^2 ds/dt = 12t but you only need 4t so it becomes 1/3 of ds/dt so $\int\limits \frac{\frac{1}{3} \times 12t}{1 + 6t^2} dt = \frac{1}{3} \int\limits \frac{4t}{1 + 6t^2} dt$

7. Outkast3r09

the 1/8 might be a typo

8. MarcLeclair

Thanks to both of you :) help is greatly appreciated haha. And it looks like it. 1/3 does seem right!