Hi,
I am having trouble at transforming integrands. I just need these to be transformed then I think I will be able to integrate it. I hope you can help me.
Thank you.
For the first, long divide.
For the second
$\displaystyle \displaystyle \begin{align*} \int{\frac{2t - 3}{\sqrt{4t - 1}}\,dt} &= \frac{1}{4}\int{\frac{2t-3}{\sqrt{4t-1}}\,4\,dt} \end{align*} $
Make the substitution $\displaystyle \displaystyle \begin{align*} u = 4t - 1 \implies du = 4\,dt \end{align*} $ and the integral becomes
$\displaystyle \displaystyle \begin{align*} \frac{1}{4}\int{\frac{2t-3}{\sqrt{4t-1}}\,4\,dt} &= \frac{1}{4}\int{\frac{\frac{1}{2}\left(u + 1\right) - 3}{\sqrt{u}}\,du} \\ &= \frac{1}{8}\int{\frac{u + 1 - 6}{\sqrt{u}}\,du} \\ &= \frac{1}{8}\int{\frac{u - 5}{\sqrt{u}}\,du} \\ &= \frac{1}{8}\int{u^{\frac{1}{2}} - 5u^{-\frac{1}{2}}\,du} \end{align*} $
Go from here.