Science:Math Exam Resources/Courses/MATH101/April 2018/Question 10 (iii)/Solution 1

Since $c$ is given by $2$ in part (ii), it is enough to evaluate the integral $$\underset{0}{\overset{1}{\int }}\frac{[2x-\log (1+2x)]}{x^2}dx.$$

We apply the integration by parts for $u(x)=2x-\log (1+2x)$ and $v^{\prime }(x)=\frac{1}{x^2}$. Then, $u^{\prime }(x)=2-\frac{2}{1+2x}=\frac{4x}{1+2x}$ , $v(x)=-\frac{1}{x}$, and $$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\frac{[2x-\log (1+2x)]}{x^2}dx& =\underset{t\to 0}{\lim }\underset{t}{\overset{1}{\int }}\frac{[2x-\log (1+2x)]}{x^2}dx\\ & =\underset{t\to 0}{\lim }\underset{t}{\overset{1}{\int }}u(x)v^{\prime }(x)dx=\underset{t\to 0}{\lim }u(x)v(x){|}_t^1-\underset{0}{\overset{1}{\int }}{u}^{\prime }(x)v(x)dx\\ & =\underset{t\to 0}{\lim }(2x-\log (1+2x))\cdot (-\frac{1}{x}){|}_t^1-\underset{0}{\overset{1}{\int }}\frac{4x}{1+2x}\cdot (-\frac{1}{x})dx\\ & =-2+\log 3+\underset{t\to 0}{\lim }\frac{(2t-\log (1+2t))}{t}+\underset{0}{\overset{1}{\int }}\frac{4}{1+2x}dx.\end{array}$$

By L'hospital rule, we can compute the limit as $$\underset{t\to 0}{\lim }\frac{(2t-\log (1+2t))}{t}=\underset{t\to 0}{\lim }\frac{(2t-\log (1+2t){)}^{\prime }}{(t{)}^{\prime }}=\underset{t\to 0}{\lim }\frac{2-\frac{2}{1+2t}}{1}=0.$$

On the other hand, using the substitution $u=1+2x$ (so $du=2dx$), the last term can be evaluated; $$\underset{0}{\overset{1}{\int }}\frac{4}{1+2x}dx=\underset{1}{\overset{3}{\int }}\frac{2}{u}du=2\log u{|}_1^3=2\log 3.$$

Combining all the information, the integral $I$ for $c=2$ has the value $$I=-2+\log 3+\underset{t\to 0}{\lim }\frac{(2t-\log (1+2t))}{t}+\underset{0}{\overset{1}{\int }}\frac{4}{1+2x}dx=-2+\log 3+0+2\log 3=\log (27)-2.$$

Answer: $\log (27)-2$