Question

Evaluate $\int e^3\log x{(x^4+1)}^{-1}dx$

• A. $e^{3\log x}+c$
• B. $\left(\frac{1}{4}\right)\log \left|x^4+1\right|+c$
• C. $\log (x^4+1)+c$
• D. $\frac{1}{2}\log (x^4+1)+c$
• E. $\left(\frac{x^4}{x^4+1}\right)+c$

Answer 1

The correct option is B

$\left(\frac{1}{4}\right)\log \left|x^4+1\right|+c$

Explanation for the correct option.

Finding the integral.

Given, $\int e^3\log x{(x^4+1)}^{-1}dx$

Let, $x^4+1=t\Rightarrow 4x^{3}dx=dt$

Now,

$$\begin{gathered} {\mathrm{e}}^{3\log \mathrm{x}}{({\mathrm{x}}^4+1)}^{-1} \\ ={\mathrm{e}}^{{\mathrm{logx}}^3}{({\mathrm{x}}^4+1)}^{-1} \\ =\frac{{\mathrm{x}}^3}{({\mathrm{x}}^4+1)}\left[\because {\mathrm{e}}^{\log \mathrm{x}}=\mathrm{x}\right] \end{gathered}$$

Substituting $t$ in the integral we get,
$$\begin{gathered} \Rightarrow \int {\mathrm{e}}^{3\log \mathrm{x}}{({\mathrm{x}}^4+1)}^{-1}d\mathrm{x}=\int \frac{{\mathrm{x}}^3}{({\mathrm{x}}^4+1)}d\mathrm{x} \\ =\frac{1}{4}\int \mathrm{t}d\mathrm{t} \\ =\frac{1}{4}\log \left|\mathrm{t}\right|+\mathrm{C} \\ =\frac{1}{4}\log \mid {\mathrm{x}}^4+1\mid +\mathrm{C}\left[\mathrm{t}={\mathrm{x}}^4+1\right] \end{gathered}$$

Hence, option (B) is correct.