Question

The value of ${\int }_2^3\frac{x+1}{x^2(x-1)}dx$ is

• A. $\log \left(\frac{16}{9}\right)+\frac{1}{6}$
• B. $\log \left(\frac{16}{9}\right)-\frac{1}{6}$
• C. $2\log \left(2\right)-\frac{1}{6}$
• D. $\log \left(\frac{4}{3}\right)-\frac{1}{6}$

Answer 1

The correct option is B

$\log \left(\frac{16}{9}\right)-\frac{1}{6}$

Explanation for correct option:

Compute the required value:

Given: ${\int }_2^3\frac{x+1}{x^2(x-1)}dx$

Using the fraction decomposition method

Let $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}=\frac{x+1}{x^2(x-1)}$

$$\begin{gathered} \Rightarrow \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}=\frac{x+1}{x^2(x-1)} \\ \Rightarrow \frac{A\left(x\right)(x-1)+B(x-1)+C\left(x^2\right)}{x^2(x-1)}=\frac{x+1}{x^2(x-1)} \\ \Rightarrow \frac{A{x}^2-Ax+Bx-B+C{x}^2}{x^2(x-1)}=\frac{x+1}{x^2(x-1)} \\ \Rightarrow \frac{x^2(A+C)+x(-A+B)-B}{x^2(x-1)}=\frac{x+1}{x^2(x-1)} \end{gathered}$$

Comparing like terms from both sides

$$\begin{gathered} -B=1 \\ \Rightarrow B=-1 \\ -A+B=1 \\ \Rightarrow A=-2 \\ A+C=0 \\ \Rightarrow C=2 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\int }_2^3\frac{x+1}{x^2(x-1)}dx={\int }_2^3\frac{-2}{x}dx+{\int }_2^3\frac{-1}{x^2}dx+{\int }_2^3\frac{2}{(x-1)}dx \\ \Rightarrow {\int }_2^3\frac{x+1}{x^2(x-1)}dx=-2{\left|\log \left(x\right)\right|}_2^3-{\left|\frac{x^{-2+1}}{-2+1}\right|}_2^3+2{\left|\mathrm{lo}g(x-1)\right|}_2^3\mathbf{\left[}\mathbf{\int }\frac{\mathbf{1}}{\mathbf{x}}\mathbf{.}\mathbf{dx}\mathbf{=}\mathbf{log}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{,}\mathbf{}\mathbf{\int }{\mathbf{x}}^{\mathbf{n}}\mathbf{.}\mathbf{d}\mathbf{x}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right]} \\ \Rightarrow {\int }_2^3\frac{x+1}{x^2(x-1)}dx=-2\left|\log \left(3\right)-\log \left(2\right)\right|-\left|\frac{1}{3}-\frac{1}{2}\right|+2\left|\log (3-1)-\log (2-1)\right| \\ \Rightarrow {\int }_2^3\frac{x+1}{x^2(x-1)}dx=-2\left|\log \left(\frac{3}{2}\right)\right|-\left|\frac{1}{6}\right|+2\left|\log \left(2\right)\right|\left[\log \left(a\right)-\log \left(b\right)=\log \left(\frac{a}{b}\right)\right] \\ \Rightarrow {\int }_2^3\frac{x+1}{x^2(x-1)}dx=\left|\log \left(\frac{4}{9}\right)\right|+\left|\log \left(4\right)\right|-\frac{1}{6} \\ \Rightarrow {\int }_2^3\frac{x+1}{x^2(x-1)}dx=\log \left(\frac{16}{9}\right)-\frac{1}{6}\left[\log \left(m\right)+\log \left(n\right)=\log \left(\mathrm{mn}\right)\right] \end{gathered}$$

Hence, option (B) is the correct answer.