- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 3,177

Compute

$\displaystyle\int_0^\frac{1}{2}\frac{4}{1-4t^2}dt$

Thot I could solve by using the log rule for integration namely:

$\displaystyle\int\frac{u'}{u} = ln|u|+C$

for such $u=1-4t^2$ and $u'=-8t$

but then $\displaystyle\frac{1}{2}\int_0^\frac{1}{2}\frac{-8t}{1-4t^2}dt$ isn't going to work due to the t in the numerator....so what shud I do....

$\displaystyle\int_0^\frac{1}{2}\frac{4}{1-4t^2}dt$

Thot I could solve by using the log rule for integration namely:

$\displaystyle\int\frac{u'}{u} = ln|u|+C$

for such $u=1-4t^2$ and $u'=-8t$

but then $\displaystyle\frac{1}{2}\int_0^\frac{1}{2}\frac{-8t}{1-4t^2}dt$ isn't going to work due to the t in the numerator....so what shud I do....

Last edited: