g. use the substitution ( u = 5x + 4 ) to evaluate the integral ( int \frac{10x + 11}{sqrt{5x + 4}} dx ).

Explanation:

Step1: Find (dx) in terms of (du)

Given (u = 5x+4), then (du=5dx), so (dx=\frac{1}{5}du). Also, (x=\frac{u - 4}{5}).

Step2: Substitute (x) and (dx) into the integral

[ \begin{align*} \int\frac{10x + 11}{\sqrt{5x+4}}dx&=\int\frac{10\times\frac{u - 4}{5}+11}{\sqrt{u}}\times\frac{1}{5}du\ &=\int\frac{2(u - 4)+11}{5\sqrt{u}}du\ &=\int\frac{2u-8 + 11}{5\sqrt{u}}du\ &=\int\frac{2u+3}{5\sqrt{u}}du\ &=\frac{1}{5}\int\left(2u^{\frac{1}{2}}+3u^{-\frac{1}{2}}\right)du \end{align*} ]

Step3: Integrate term - by - term

Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have: [ \begin{align*} \frac{1}{5}\int\left(2u^{\frac{1}{2}}+3u^{-\frac{1}{2}}\right)du&=\frac{1}{5}\left(2\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+3\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right)+C\ &=\frac{1}{5}\left(\frac{4}{3}u^{\frac{3}{2}}+6u^{\frac{1}{2}}\right)+C \end{align*} ]

Step4: Substitute back (u = 5x+4)

[ \begin{align*} \frac{1}{5}\left(\frac{4}{3}(5x + 4)^{\frac{3}{2}}+6(5x + 4)^{\frac{1}{2}}\right)+C&=\frac{4}{15}(5x + 4)^{\frac{3}{2}}+\frac{6}{5}(5x + 4)^{\frac{1}{2}}+C\ &=\frac{2}{15}(5x + 4)^{\frac{1}{2}}(2(5x + 4)+18)+C\ &=\frac{2}{15}(5x + 4)^{\frac{1}{2}}(10x+8 + 18)+C\ &=\frac{2}{15}(5x + 4)^{\frac{1}{2}}(10x + 26)+C \end{align*} ]

Answer:

(\frac{4}{15}(5x + 4)^{\frac{3}{2}}+\frac{6}{5}(5x + 4)^{\frac{1}{2}}+C)