hi there,
could you help me solve this question out?
Mr. shankar venkatagiri?
 
the question is
 
integral of 1/1+x^4 dx ?

Asked by ankit malhotra | 29th Sep, 2014, 10:00: PM

Expert Answer:

C o n s i d e r space t h e space e v a l u a t i o n space o f space t h e space i n t e g r a l comma space integral fraction numerator d x over denominator 1 plus x to the power of 4 end fraction : I equals integral fraction numerator d x over denominator 1 plus x to the power of 4 end fraction equals 1 half integral fraction numerator 1 plus x squared plus 1 minus x squared over denominator 1 plus x to the power of 4 end fraction d x equals 1 half open square brackets integral fraction numerator 1 plus x squared over denominator 1 plus x to the power of 4 end fraction d x plus integral fraction numerator 1 minus x squared over denominator 1 plus x to the power of 4 end fraction d x close square brackets equals 1 half open square brackets integral fraction numerator 1 plus begin display style 1 over x squared end style over denominator x squared plus 1 over x squared end fraction d x minus integral fraction numerator 1 minus begin display style 1 over x squared end style over denominator x squared plus 1 over x squared end fraction d x close square brackets equals 1 half open square brackets integral fraction numerator 1 plus begin display style 1 over x squared end style over denominator open parentheses x minus begin display style 1 over x end style close parentheses squared plus 2 end fraction d x minus integral fraction numerator 1 minus begin display style 1 over x squared end style over denominator open parentheses x plus 1 over x close parentheses squared minus 2 end fraction d x close square brackets N o w space s u b s t i t u t e space x minus 1 over x equals t space a n d space x plus 1 over x equals u ; rightwards double arrow open parentheses 1 plus 1 over x squared close parentheses d x equals d t space a n d space open parentheses 1 minus 1 over x squared close parentheses d x equals d u rightwards double arrow I equals 1 half open square brackets integral fraction numerator begin display style 1 end style over denominator t squared plus 2 end fraction d x minus integral fraction numerator begin display style 1 end style over denominator u squared minus 2 end fraction d u close square brackets rightwards double arrow I equals 1 half open square brackets integral fraction numerator begin display style 1 end style over denominator t squared plus 2 end fraction d x minus integral fraction numerator begin display style 1 end style over denominator u squared minus 2 end fraction d u close square brackets rightwards double arrow I equals 1 half open square brackets fraction numerator 1 over denominator square root of 2 end fraction tan to the power of minus 1 end exponent fraction numerator t over denominator square root of 2 end fraction minus fraction numerator 1 over denominator 2 square root of 2 end fraction log fraction numerator u minus square root of 2 over denominator u plus square root of 2 end fraction close square brackets rightwards double arrow I equals 1 half open square brackets fraction numerator 1 over denominator square root of 2 end fraction tan to the power of minus 1 end exponent fraction numerator open parentheses x minus 1 over x close parentheses over denominator square root of 2 end fraction minus fraction numerator 1 over denominator 2 square root of 2 end fraction log fraction numerator space x plus 1 over x minus square root of 2 over denominator space x plus 1 over x plus square root of 2 end fraction close square brackets

Answered by Vimala Ramamurthy | 30th Sep, 2014, 11:36: AM