Answer to Question #314792 in Real Analysis for Jyo

Question

Apply second substitution theorem evaluate

i) integral 1 to 9 (√ t)/(2+√t)

Answer 1

$\int \frac{\sqrt{t}}{\sqrt{t}+2} \mathrm{~d} t\\[2mm] \text{Substitute } u=\sqrt{t}+2 \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} t}=\frac{1}{2 \sqrt{t}}\longrightarrow \mathrm{d} t=2 \sqrt{t} \mathrm{~d} u, \text{ use:}\\ \begin{aligned} & t=(u-2)^{2} \\ =& 2 \int \frac{(u-2)^{2}}{u} \mathrm{~d} u \end{aligned} \\[4mm] \text{Now solving:}\\[2mm] \begin{aligned} & \int \frac{(u-2)^{2}}{u} \mathrm{~d} u \\=& \int\left(u+\frac{4}{u}-4\right) \mathrm{d} u \end{aligned}\\[2mm] \text{Apply linearity:}\\[2mm] =\int u \mathrm{~d} u+4 \int \frac{1}{u} \mathrm{~d} u-4 \int 1 \mathrm{~d} u\\[3mm] \text{Now solving:}\\[2mm] \int u \mathrm{~d} u\\[2mm] \text{Apply power rule:}\\[2mm] \begin{aligned} \int u^{\mathrm{n}} \mathrm{d} u=& \frac{u^{\mathrm{n}+1}}{\mathrm{n}+1} \text { with } \mathrm{n}=1 \\ &=\frac{u^{2}}{2} \end{aligned} \\[2mm]$

$\text{Now solving:}\\[2mm] \int \frac{1}{u} \mathrm{~d} u\\[2mm] \text{This is a standard integral:}\\[2mm] =\ln (u)\\[3mm] \text{Now solving:}\\[2mm] \int 1 \mathrm{~d} u\\[2mm] \text{Apply constant rule:}\\[2mm] =\boldsymbol{u}\\[2mm] \text{Plug in solved integrals:}\\[2mm] \begin{gathered} \int u \mathrm{~d} u+4 \int \frac{1}{u} \mathrm{~d} u-4 \int 1 \mathrm{~d} u \\ =4 \ln (u)+\frac{u^{2}}{2}-4 u \end{gathered}\\[2mm] \text{Plug in solved integrals:}\\[2mm] 2 \int \frac{(u-2)^{2}}{u} \mathrm{~d} u\\=8 \ln (u)+u^{2}-8 u\\[2mm] \text{ Undo substitution }\\ u=\sqrt{t}+2 =-8(\sqrt{t}+2)+(\sqrt{t}+2)^{2}+8 \ln (\sqrt{t}+2)\\[2mm] \text{The problem is solved:}\\[2mm] \begin{gathered} \int \frac{\sqrt{t}}{\sqrt{t}+2} \mathrm{~d} t \\ =-8(\sqrt{t}+2)+(\sqrt{t}+2)^{2}+8 \ln (\sqrt{t}+2)+C \end{gathered}$ $=t-4 \sqrt{t}+8 \ln (\sqrt{t}+2)+C$

Evaluating the above function at [1,9], we have:

$8 \ln (5)-8 \ln (3)\\[3mm] \text{Simplify/rewrite:}\\[2mm] 8(\ln (5)-\ln (3))\\[3mm] \text{ Approximation:}\\4.086604990127925$