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Calculus Level 4

$\large \sqrt2 \int \dfrac{(x-1)(x+1)}{\sqrt{(x^4+1)(x^2+1)^2}} \, dx = \, ?$

Notation : $C$ denotes the constant of arbitration.

\dfrac1{\sqrt2} \arctan\sqrt{x^2+1}+C

\arctan\sqrt{x^2+\dfrac1{x^2}}+C

\arctan\sqrt{ \dfrac12 \left(x^2+\dfrac1{x^2}\right)}+C

\dfrac12\arctan\sqrt{x^2+\dfrac1{x^2}}+C

1 solution

Mohammad Hamdar
Jan 12, 2016

$The\quad integral\quad may\quad be\quad written\quad as\int \frac { x^{ 2 }-1 }{ x(x^{ 2 }+1)\sqrt { x^{ 2 }+\frac { 1 }{ { x }^{ 2 } } } } \, dx\\ =\int \frac { 1-\frac { 1 }{ { x }^{ 2 } } }{ (x+\frac { 1 }{ x } )\sqrt { (x+\frac { 1 }{ x } )^{ 2 }-2 } } \, dx\\ Use\quad the\quad substitution\quad t=x+\frac { 1 }{ x } \\ which\quad transforms\quad the\quad integral\quad into\int \frac { 1 }{ t\sqrt { t^{ 2 }-2 } } \, dt.\\ let\quad t=\sqrt { 2 } \sec { \theta } \\ which\quad transforms\quad the\quad integral\quad into\int \frac { 1 }{ \sqrt { 2 } } \, d\theta =\frac { \theta }{ \sqrt { 2 } } +c\\ which\quad equals\quad \frac { 1 }{ \sqrt { 2 } } \arctan \sqrt { \frac { 1 }{ 2 } (x^{ 2 }+\frac { 1 }{ x^{ 2 } } ) } +C.$

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