Question

The value of ${\int }_0^1\frac{x^4+1}{x^2+1}dx$

• A. $\frac{1}{6}(3–4\pi )$
• B. $\frac{1}{6}(3\mathrm{\pi }+4)$
• C. $\frac{1}{6}(3+4\pi )$
• D. $\frac{1}{6}(3\mathrm{\pi }–4)$

Answer 1

The correct option is D

$\frac{1}{6}(3\mathrm{\pi }–4)$

The explanation for the correct answer.

Solve for the value of integral.

$$\begin{gathered} I={\int }_0^1\frac{x^4+1}{x^2+1}dx \\ ={\int }_0^1\frac{{\left(x^2+1\right)}^2}{x^2+1}-\frac{2x^2}{x^2+1}dx \\ ={\int }_0^1\left(x^2+1\right)dx-2{\int }_0^1\frac{x^2}{x^2+1}dx \\ ={\int }_0^1\left(x^2+1\right)dx-2{\int }_0^1\frac{x^2+1-1}{x^2+1}dx \\ ={\int }_0^1\left(x^2+1\right)dx-2{\int }_0^1\frac{x^2+1}{x^2+1}dx+2{\int }_0^1\frac{1}{x^2+1}dx \\ ={\left[\frac{x^3}{3}+x\right]}_0^1-2{\left[x\right]}_0^1+2{\left[{\tan }^{-1}x\right]}_0^1 \\ =\left(\frac{1}{3}+1\right)-0-2\left(1-0\right)+2\left(\frac{\mathrm{\pi }}{4}-0\right) \\ =\frac{4}{3}-2+\frac{\mathrm{\pi }}{2} \\ =\frac{\mathrm{\pi }}{2}-\frac{2}{3} \\ =\frac{3\mathrm{\pi }-4}{6} \end{gathered}$$

Hence, option(D) is the correct answer.