Math, asked by Anonymous, 1 year ago

integration of x^2/1+x^4

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Answers

Answered by 02020divaynshu

I = ∫ [ x² / ( x⁴ + 1 ) ] dx

= (1/2) ∫ [ ( 2x² ) / ( x⁴ + 1 ) ] dx

= (1/2) ∫ { [ ( x² + 1 ) + ( x² - 1 ) ] / ( x⁴ + 1 ) } dx

= (1/2) ∫ [ ( x² + 1 ) / ( x⁴ + 1 ) ] dx + (1/2) ∫ [ ( x² - 1 ) / ( x⁴ + 1 ) ] dx
..........................................

In each integrand, dividing Num and Den by x², we get
..........................................

= (1/2) ∫ { [ 1 + (1/x²) ] / [ x² + (1/x²) ] } dx + (1/2) ∫ { [ 1 - (1/x²) ] / [ x² + (1/x²) ] } dx

= (1/2) ∫ {[ 1-(-1/x²) ] / [ (x- 1/x)² + 2 ]} dx + (1/2) ∫ {[ 1+(-1/x²) ] / [ (x+ 1/x)² - 2 ]} dx

= (1/2) ∫ { 1 / [ u²+(√2)² ) ]} du + (1/2) ∫ { 1 / [ v²-(√2)² ]} dv, ... u = x - 1/x, v = x + 1/x

= (1/2) • (1/√2) tanֿ¹ (u/√2) + (1/2) • [1/(2√2)] ln | ( v - √2 ) / ( v + √2 ) | + C

= [1/(2√2)]·tanֿ¹ [ (x²-1) / (x√2) ] + [1/(4√2)]·ln |(x² -√2.x + 1) / (x² +√2.x + 1)| + C.

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Answered by ujjwalbittuyaar

Answer:

I = ∫ [ x² / ( x⁴ + 1 ) ] dx

= (1/2) ∫ [ ( 2x² ) / ( x⁴ + 1 ) ] dx

= (1/2) ∫ { [ ( x² + 1 ) + ( x² - 1 ) ] / ( x⁴ + 1 ) } dx

= (1/2) ∫ [ ( x² + 1 ) / ( x⁴ + 1 ) ] dx + (1/2) ∫ [ ( x² - 1 ) / ( x⁴ + 1 ) ] dx

..........................................

In each integrand, dividing Num and Den by x², we get

..........................................

= (1/2) ∫ { [ 1 + (1/x²) ] / [ x² + (1/x²) ] } dx + (1/2) ∫ { [ 1 - (1/x²) ] / [ x² + (1/x²) ] } dx

= (1/2) ∫ {[ 1-(-1/x²) ] / [ (x- 1/x)² + 2 ]} dx + (1/2) ∫ {[ 1+(-1/x²) ] / [ (x+ 1/x)² - 2 ]} dx

= (1/2) ∫ { 1 / [ u²+(√2)² ) ]} du + (1/2) ∫ { 1 / [ v²-(√2)² ]} dv, ... u = x - 1/x, v = x + 1/x

= (1/2) • (1/√2) tanֿ¹ (u/√2) + (1/2) • [1/(2√2)] ln | ( v - √2 ) / ( v + √2 ) | + C

= [1/(2√2)]·tanֿ¹ [ (x²-1) / (x√2) ] + [1/(4√2)]·ln |(x² -√2.x + 1) / (x² +√2.x + 1)| + C.

Step-by-step explanation:

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