Integral Help

[Badger2014]

\(\displaystyle \displaystyle x\, =\, 75\, \mbox{exp}\, \left(\int_0^3\, \delta t\, dt\right)\,\)

. .\(\displaystyle \displaystyle=\, 75\, \mbox{exp}\, \left(\int_0^3\, \dfrac{2}{k\, +\, 2t}\, dt\right)\)

. .\(\displaystyle =\, 75\, \mbox{exp}\, \left(\, \ln\left(2t\, +\, k\right)\bigg\rvert_0^3\, \right)\)

. .\(\displaystyle =\, 75\, \dfrac{(6\, +\, k)}{k}\)

Can someone please explain how they get (ln(2t+k))? And then how it simplifies to ((6+k)/k)? I know the e^(ln(x)) rule leads to x and I know that the 3 from the upper bound is plugged in for t to get the 6 but I don't know how the k ends up in the denominator?

Thanks

 

[Deleted member 4993]

\(\displaystyle \displaystyle x\, =\, 75\, \mbox{exp}\, \left(\int_0^3\, \delta t\, dt\right)\,\)

. .\(\displaystyle \displaystyle=\, 75\, \mbox{exp}\, \left(\int_0^3\, \dfrac{2}{k\, +\, 2t}\, dt\right)\)

. .\(\displaystyle =\, 75\, \mbox{exp}\, \left(\, \ln\left(2t\, +\, k\right)\bigg\rvert_0^3\, \right)\)

. .\(\displaystyle =\, 75\, \dfrac{(6\, +\, k)}{k}\)

Can someone please explain how they get (ln(2t+k))? And then how it simplifies to ((6+k)/k)? I know the e^(ln(x)) rule leads to x and I know that the 3 from the upper bound is plugged in for t to get the 6 but I don't know how the k ends up in the denominator?

What would you get for:

\(\displaystyle \displaystyle{\frac{d}{dt}\left[ln(2t+k)\right ] \ = \ ??}\)

Now compare that with the integrand!!