Question

The value of ${\int }_1^2\frac{1}{\left[x(1+x^4)\right]}dx$ is

• A. $\frac{1}{4}\log \left(\frac{17}{32}\right)$
• B. $\frac{1}{4}\log \left(\frac{32}{17}\right)$
• C. $\log \left(\frac{17}{2}\right)$
• D. $\frac{1}{4}\log \left(\frac{17}{2}\right)$

Answer 1

The correct option is B

$\frac{1}{4}\log \left(\frac{32}{17}\right)$

Explanation for correct option:

Compute the required value:

Given: ${\int }_1^2\frac{1}{\left[x(1+x^4)\right]}dx$

multiply $x^3$to both numerator and denominator

${\int }_1^2\frac{x^3}{\left[x^4(1+x^4)\right]}dx$

Let $x^4=t\Rightarrow 4x^{3}dx=dt$

When $$\begin{gathered} x=1,t=1 \\ x=2,t=16 \end{gathered}$$

$$\begin{gathered} =\frac{1}{4}{\int }_1^{16}\frac{1}{\left[t(1+t)\right]}dt \\ =\frac{1}{4}{\int }_1^{16}\frac{t+1-t}{\left[t(1+t)\right]}dt \\ =\frac{1}{4}{\int }_1^{16}\left(\frac{1}{t}-\frac{1}{(1+t)}\right)dt \\ =\frac{1}{4}{\int }_1^{16}\left(\frac{1}{t}\right)·dt-{\int }_1^{16}\left(\frac{1}{(1+t)}\right)dt \\ =\frac{1}{4}{\left|\log \left(t\right)\right|}_1^{16}-\frac{1}{4}{\left|\log (1+t)\right|}_1^{16}\left[as\mathbf{\int }\frac{\mathbf{1}}{\mathbf{x}}\mathbf{d}\mathit{x}\mathbf{=}\mathit{l}\mathit{o}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\right] \\ =\frac{1}{4}\left[\log \left(16\right)-\log \left(1\right)-\log \left(17\right)+\log \left(2\right)\right] \\ =\frac{1}{4}\left[4\log \left(2\right)+\log \left(2\right)-\log \left(17\right)\right] \\ =\frac{1}{4}\left[5\log \left(2\right)-\log \left(17\right)\right] \\ =\frac{1}{4}\left[\log \left(32\right)-\log \left(17\right)\right] \\ =\frac{1}{4}\left[\log \left(\frac{32}{17}\right)\right] \end{gathered}$$

Hence, option $B$ is the correct answer.