# evaluate the integral (dt)/(3-4t) the lower limit is 5 and upper is 2

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move from 1st to second by multiplying denominator by (-1).

From 2nd to 3rd swap lower and upper bound by multiplying the integral by (-)

next step we know that int(1/x)=lnx. Since our problem has 4t-3, we have to make up for it by dividing by 4.

Hope this clarify the work.

`int_5^2dt/(3-4t)=`

`-int_5^2dt/(4t-3)=`

` ` `int_2^5dt/(4t-3)=`

`1/4*ln(4t-3) between 2 and 5=`

`1/4(ln5-ln17)`