Now we need to get set up for the substitution. In other words, we need so solve for $t$ and get $dt$.

$$t = \frac{1}{2}{u^2} + 2\hspace{0.5in} \Rightarrow \hspace{0.5in}dt = u\,du$$ Show Step 3

Doing the substitution gives,

$$\int{{\frac{{t - 2}}{{t - 3\sqrt {2t - 4} + 2}}\,dt}} = \int{{\frac{{\frac{1}{2}{u^2} + 2 - 2}}{{\frac{1}{2}{u^2} + 2 - 3u + 2}}\,\left( u \right)du}} = \int{{\frac{{{u^3}}}{{{u^2} - 6u + 8}}\,du}}$$ Show Step 4

This integral requires partial fractions to evaluate.

However, we first need to do long division on the integrand since the degree of the numerator (3) is higher than the degree of the denominator (2). This gives,

$$\frac{{{u^3}}}{{{u^2} - 6u + 8}} = u + 6 + \frac{{28u - 48}}{{\left( {u - 2} \right)\left( {u - 4} \right)}}$$

The form of the partial fraction decomposition on the third term is,

$$\frac{{28u - 48}}{{\left( {u - 2} \right)\left( {u - 4} \right)}} = \frac{A}{{u - 2}} + \frac{B}{{u - 4}}$$

Setting the coefficients equal gives,

$$28u - 48 = A\left( {u - 4} \right) + B\left( {u - 2} \right)$$

Using the “trick” to get the coefficients gives,

$$\begin{align*}{u = 4:} & \hspace{0.25in}& 64 = & \, 2B\\{u = 2:} & \hspace{0.25in} & 8 = & - 2A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in} \begin{aligned} & {A = - 4}\\ & {B = 32}\end{aligned}$$

The integral is then,

$$\int{{\frac{{{u^3}}}{{{u^2} - 6u + 8}}\,du}} = \int{{u + 6 - \frac{4}{{u - 2}} + \frac{{32}}{{u - 4}}\,du}} = \frac{1}{2}{u^2} + 6u - 4\ln \left| {u - 2} \right| + 32\ln \left| {u - 4} \right| + c$$ Show Step 5

The last step is to now do all the back substitutions to get the final answer.

$$\int{{\frac{{{u^3}}}{{{u^2} - 6u + 8}}\,du}} = \require{bbox} \bbox[2pt,border:1px solid black]{{t - 2 + 6\sqrt {2t - 4} - 4\ln \left| {\sqrt {2t - 4} - 2} \right| + 32\ln \left| {\sqrt {2t - 4} - 4} \right| + c}}$$