How do you find the integral of int 1/(4y-1) dy∫14y−1dy from 0 to 1?

1 Answer
Sasha P.
Sep 26, 2015

ln3/4ln34

Explanation:

4y-1=t => 4dy=dt => dy=dt/44y−1=t⇒4dy=dt⇒dy=dt4
t_1=4*0-1=-1t1=4⋅0−1=−1
t_2=4*1-1=3t2=4⋅1−1=3

I=int_0^1 1/(4y-1)dy = int_-1^3 1/t dt/4 = 1/4 int_-1^3 dt/t = 1/4 ln|t| |_-1^3I=∫1014y−1dy=∫3−11tdt4=14∫3−1dtt=14ln|t|∣∣∣3−1

I=1/4(ln|3|-ln|-1|)=1/4(ln3-ln1)=1/4(ln3-0)I=14(ln|3|−ln|−1|)=14(ln3−ln1)=14(ln3−0)

I=ln3/4I=ln34

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