Mathematics

# If $\int { \cfrac { xdx }{ \sqrt { 1+{ x }^{ 2 }+\sqrt { { \left( 1+{ x }^{ 2 } \right) }^{ 3 } } } } } =k\sqrt { 1+\sqrt { 1+{ x }^{ 2 } } } +C$ then k equals to

##### SOLUTION
Put, $1+{ x }^{ 2 }={ z }^{ 2 }$
$2zdz=2xdx$
$\int { \cfrac { xdx }{ \sqrt { 1+{ x }^{ 2 }+\sqrt { { \left( 1+{ x }^{ 2 } \right) }^{ 3 } } } } }$
$=\int { \cfrac { zdz }{ \sqrt { { z }^{ 2 }+{ z }^{ 3 } } } }$
$=\int { \cfrac { zdz }{ z\sqrt { 1+{ z } } } } =\int { \cfrac { dz }{ \sqrt { 1+{ z } } } } =\int { { \left( 1+z \right) }^{ -{ 1 }/{ 2 } }\cdot dz }$
$=\cfrac { 1 }{ -{ 1 }/{ 2 }+1 } \times { \left( 1+z \right) }^{ -{ 1 }/{ 2 }+1 }+C$
$=\cfrac { 1 }{ 1/2 } { (1+z) }^{ { 1 }/{ 2 } }+C$
$2\sqrt { 1+z } +C$
$=2\sqrt { 1+\sqrt { 1+{ x }^{ 2 } } } +C$
Ans: k=2

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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