Question

If $\int_{}^{}\frac{1}{x^{4} + 1}$dx = $\frac{1}{2\sqrt{2}}$tan^–1$\left( \frac{x^{2} - 1}{\sqrt{2}x} \right)$ + A + c. Then A is equal to –

• A. – $\frac{1}{2\sqrt{2}}$lo$\left| \frac{x^{2} - \sqrt{2}x + 1}{x^{2} + \sqrt{2}x + 1} \right|$
• B. – $\frac{1}{4\sqrt{2}}$log $\left| \frac{x^{2} - \sqrt{2}x + 1}{x^{2} + \sqrt{2}x + 1} \right|$
• C. –$\frac{1}{2}$log$\left| \frac{x^{2} + \sqrt{2}x + 1}{x^{2}–\sqrt{2}x + 1} \right|$
• D. – $\frac{1}{4\sqrt{2}}$log $\left| \frac{x^{2} + \sqrt{2}x + 1}{x^{2}–\sqrt{2}x + 1} \right|$

Answer 1

Answer: (B)

– $\frac{1}{4\sqrt{2}}$log $\left| \frac{x^{2} - \sqrt{2}x + 1}{x^{2} + \sqrt{2}x + 1} \right|$

Answer 2

$\int \frac { 1 } { x ^ { 4 } + 1 }$dx

= $\int \frac { 1 / x ^ { 2 } } { x ^ { 2 } + 1 / x ^ { 2 } }$ dx

=$\frac { 1 } { 2 }$ $\int_{}^{}\frac{2/x^{2}}{x^{2} + 1/x^{2}}$dx

=dx

= $\frac { 1 } { 2 } \int \frac { 1 + \frac { 1 } { x ^ { 2 } } } { x ^ { 2 } + \frac { 1 } { x ^ { 2 } } }$ dx – $\frac{1}{2}\ \int_{}^{}\frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}}$dx

= $\frac { 1 } { 2 }$ $\int_{}^{}\frac{1}{\left( x - \frac{1}{x} \right)^{2} + 2}$d$\left( x - \frac{1}{x} \right)$

– $\frac { 1 } { 2 }$ $\int_{}^{}\frac{1}{\left( x + \frac{1}{x} \right)^{2} - 2}$d $\left( x + \frac{1}{x} \right)$

=$\frac { 1 } { 2 }$ $\int_{}^{}\frac{du}{u^{2} + (\sqrt{2})^{2}}$– $\frac{1}{2}$ $\int_{}^{}\frac{dv}{v^{2} - (\sqrt{2})^{2}}$,

Where u = x – $\frac{1}{x}$ and v = x +$\frac{1}{x}$

=$\frac{1}{2}$ × $\frac{1}{\sqrt{2}}$ tan^–1 $\left( \frac{u}{\sqrt{2}} \right)$– $\frac{1}{2}$ × $\frac{1}{2\sqrt{2}}$log $\left| \frac{v - \sqrt{2}}{v + \sqrt{2}} \right|$ + c

= $\frac{1}{2\sqrt{2}}$ tan^–1 $\left( \frac{x^{2} - 1}{\sqrt{2}x} \right)$ – $\frac{1}{4\sqrt{2}}$ log $\left| \frac{x^{2} - \sqrt{2}x + 1}{x^{2} + \sqrt{2}x + 1} \right|$ + c

Hence (2) is the correct answer.