Question

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.

${\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}$

Short answer

${\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}=\frac{(\ln 4)}{2}+\sum _{\mathrm{k}=1}^{\infty }\frac{{(-1)}^{\mathrm{k}+1}}{\mathrm{k}.2^{2\mathrm{k}}}\frac{(1-0.5^{2\mathrm{k}+1})}{2\mathrm{k}+1}$

Step-by-step solution

Step 1. Given information is: 

${\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From}\mathrm{Q}46. \\ \mathrm{Maclaurin}\mathrm{series}\mathrm{for}\mathrm{f}\left(\mathrm{x}\right)=\ln (4+{\mathrm{x}}^2)\mathrm{is} \\ \ln (4+{\mathrm{x}}^2)=\ln 4+\sum _{\mathrm{k}=1}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\mathrm{k}.2^{2\mathrm{k}}}.{\mathrm{x}}^{2\mathrm{k}} \\ \mathrm{Also},\mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right)=(\ln 4)\mathrm{x}+\sum _{\mathrm{k}=1}^{\infty }\frac{{(-1)}^{\mathrm{k}+1}}{\mathrm{k}.2^{2\mathrm{k}}}\frac{{\mathrm{x}}^{2\mathrm{k}+1}}{2\mathrm{k}+1} \\ \mathrm{Adding}\mathrm{the}\mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)={\left[(\ln 4)\mathrm{x}+\sum _{\mathrm{k}=1}^{\infty }\frac{{(-1)}^{\mathrm{k}+1}}{\mathrm{k}.2^{2\mathrm{k}}}\frac{{\mathrm{x}}^{2\mathrm{k}+1}}{2\mathrm{k}+1}\right]}_{0.5}^1 \\ =(\ln 4)(1-0.5)+\sum _{\mathrm{k}=1}^{\infty }\frac{{(-1)}^{\mathrm{k}+1}}{\mathrm{k}.2^{2\mathrm{k}}}\frac{(1^{2\mathrm{k}+1}-0.5^{2\mathrm{k}+1})}{2\mathrm{k}+1} \\ =\frac{(\ln 4)}{2}+\sum _{\mathrm{k}=1}^{\infty }\frac{{(-1)}^{\mathrm{k}+1}}{\mathrm{k}.2^{2\mathrm{k}}}\frac{(1-0.5^{2\mathrm{k}+1})}{2\mathrm{k}+1} \end{gathered}$$