Question

integration of x²-4/x⁴+24x²+16 dx equals​

Answer 1

Answer:

{ln(x²-12+8√2)} /4 - ( 11+5√2)  tan^-1{x/(√8 - 2)}

Step-by-step explanation:

first find the partial fractions

x²-4/x⁴+24x²+16 = A/x² + 12-8√2 + B/x² - 12+8√2

x² - 4 =  A(x² - 12+8√2) + B(x² + 12 - 8√2 )

@ x = 12-8√2  , 2B(12-8√2) = 8-8√2 , B = -(1+5√2)

@ x = 12+8√2  , 2A(12+8√2) = 8+8√2 , A = √2 - 1

so far

I = ∫x²-4/x⁴+24x²+16 dx = ∫  A/(x² - 12+8√2) .dx + ∫B/(x² + 12 - 8√2 ) dx

∫Adx/{x² - (12-8√2) } = ∫Adx/{x² - (12-2√32) }  =  ∫Adx/{x² - (√8-2)² }

= A/2(√8 -2) {ln(x²-12+8√2)} = {ln(x²-12+8√2)} /4

∫Bdx/(x² + 12 - 8√2 ) dx = B ∫dx/{x² +  (√8-2 )² } = B ∫dx/{x² +  (√8 - 2 )² }

B/(√8 - 2 ) tan^-1{x/(√8 - 2)} = - ( 11+5√2)  tan^-1{x/(√8 - 2)}

therefore

I = ∫x²-4/x⁴+24x²+16 dx = ∫  A/(x² - 12+8√2) .dx + ∫B/(x² + 12 - 8√2 ) dx

I = {ln(x²-12+8√2)} /4 - ( 11+5√2)  tan^-1{x/(√8 - 2)}