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

$$\int _{0}^{1}{\frac {\left[2x-\log(1+2x)\right]}{x^{2}}}dx.$$

We apply the integration by parts for $u(x)=2x-\log(1+2x)$ and $v'(x)={\frac {1}{x^{2}}}$. Then, $u'(x)=2-{\frac {2}{1+2x}}={\frac {4x}{1+2x}}$ , $v(x)=-{\frac {1}{x}}$, and

$${\begin{aligned}\int _{0}^{1}{\frac {\left[2x-\log(1+2x)\right]}{x^{2}}}dx&=\lim _{t\to 0}\int _{t}^{1}{\frac {\left[2x-\log(1+2x)\right]}{x^{2}}}dx\\&=\lim _{t\to 0}\int _{t}^{1}u(x)v'(x)dx=\lim _{t\to 0}u(x)v(x){\bigg |}_{t}^{1}-\int _{0}^{1}u'(x)v(x)dx\\&=\lim _{t\to 0}(2x-\log(1+2x))\cdot \left(-{\frac {1}{x}}\right){\bigg |}_{t}^{1}-\int _{0}^{1}{\frac {4x}{1+2x}}\cdot \left(-{\frac {1}{x}}\right)dx\\&=-2+\log 3+\lim _{t\to 0}{\frac {(2t-\log(1+2t))}{t}}+\int _{0}^{1}{\frac {4}{1+2x}}dx.\end{aligned}}$$

By L'hospital rule, we can compute the limit as

$$\lim _{t\to 0}{\frac {(2t-\log(1+2t))}{t}}=\lim _{t\to 0}{\frac {(2t-\log(1+2t))'}{(t)'}}=\lim _{t\to 0}{\frac {2-{\frac {2}{1+2t}}}{1}}=0.$$

On the other hand, using the substitution $u=1+2x$ (so $du=2dx$), the last term can be evaluated;

$$\int _{0}^{1}{\frac {4}{1+2x}}dx=\int _{1}^{3}{\frac {2}{u}}du=2\log u{\bigg |}_{1}^{3}=2\log 3.$$

Combining all the information, the integral $I$ for $c=2$ has the value

$$I=-2+\log 3+\lim _{t\to 0}{\frac {(2t-\log(1+2t))}{t}}+\int _{0}^{1}{\frac {4}{1+2x}}dx=-2+\log 3+0+2\log 3=\log(27)-2.$$

Answer: $\color {blue}\log(27)-2$