IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

mufeed

integral(0 to 1) (x^2)*ln(1-(x^2))dx=((k/(k+1))*ln k)-((4*k)/(3*(k+1))) where ln-natural logaritham

and (1/(1*5))+(1/(2*7))+(1/(3*9))+.....................infinity=(t/(t+1))-((3/4)*(t/(t+1))*ln(t/4))

find (k+t)

freetam

i m unable to understand this question , plz let me kno it clearly then i'll able to do it send me the same on xtreme_brett.lee@rediffmail.com but in simple form as given in books

ok

amar.gupta

Dear,

as we know : ln(1+x) =x-x²/2+x³/3-x⁴/4+......................infinity

and ln(1-x) =-x - x²/2 - x³/3 - x⁴/4 - ......................infinity

add both : ln(1+x) + ln(1-x) = -2[x²/2 + x⁴/4 + x⁶/6 ......................infinity]

or ln(1+x)(1-x) = -2[x²/2 + x⁴/4 + x⁶/6 ......................infinity]

or ln(1-x²) = -2[x²/2 + x⁴/4 + x⁶/6 ......................infinity]

or x² ln(1-x²) = -2[x⁴/2 + x⁶/4 + x⁸/6 ......................infinity]

integrate them you will get : -2[x⁵/2.5 + x⁷/4.7 + x⁹/6.9 ......................infinity]

or -[x⁵/1.5 + x⁷/2.7 + x⁹/3.9 ......................infinity]

apply limits you get : LHS= -[1/1.5 + 1/2.7 + 1/3.9 ......................infinity]

this is your given series put the value and equate with RHS and you will get the ans.

vinuthna

The question is challenging and good . It is really a good question

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