 Hi and welcome to the session, I am Poorwa and I will help you with the following question. The value of integral limit from 0 to 1 tan inverse 2x minus 1 upon 1 plus x minus x square dx is a 1 b 0 c minus 1 d pi by 4. Let us now begin with the solution. Let us denote this definite integral by i. So we have i is equal to integral limit from 0 to 1 tan inverse 2x minus 1 upon 1 plus x minus x square dx. This is equal to integral limit from 0 to 1 tan inverse. Now we can write 2x as x plus x minus 1 upon and we can write the denominator as 1 minus x into x minus 1 dx. Now since tan inverse x plus y upon 1 minus x y is equal to tan inverse x plus tan inverse y. So we can write this as this is equal to integral limit from 0 to 1 tan inverse x plus tan inverse. Now taking y as x minus 1 we get x minus 1 dx since tan inverse x plus y upon 1 minus x y is equal to tan inverse x plus tan inverse y and this is equal to integral limit from 0 to 1 1 into tan inverse x dx plus integral limit from 0 to 1 1 into tan inverse x minus 1 dx this is equal to now taking tan inverse x as the first function and 1 as the second function and applying Bipart's method we get first that is tan inverse x into integration of second so we get x minus integration limit from 0 to 1 derivative of first that is derivative of tan inverse x is 1 upon 1 plus x square into integration of second so we get x dx plus here again taking tan inverse x minus 1 as the first function and 1 as the second function and applying Bipart's method we get first that is tan inverse x minus 1 into integration of second so we get x limit is from 0 to 1 minus integration limit from 0 to 1 derivative of first that is 1 upon 1 plus x minus 1 whole square into integration of second that is x dx this is equal to now putting the limits here we get tan inverse 1 minus tan inverse 0 into 0 because we have the upper limit as 1 so putting the upper limit 1 in place of x we get tan inverse 1 into 1 minus putting lower limit 0 in place of x we get tan inverse 0 into 0 minus integral limit from 0 to 1 x upon 1 plus x square dx minus plus now again putting the limits we get tan inverse 0 minus 0 right because here the upper limit is 1 so putting 1 in place of x we get here tan inverse 0 into 1 minus putting lower limit 0 here we get this is equal to 0 minus integral limit from 0 to 1 x upon 1 plus x minus 1 whole square dx and this is equal to now tan inverse 1 is equal to pi by 4 so we have pi by 4 minus 0 minus integral limit from 0 to 1 x upon 1 plus x square dx plus now tan inverse 0 is 0 so we get 0 minus 0 which is again 0 so we get 0 minus integral limit from 0 to 1 x upon 1 plus x minus 1 whole square dx this is equal to pi by 4 minus now let us take this definite integral as I1 so we get I1 and let us take this definite integral as I2 so we get minus I2 now let us first consider I1 so we have I1 is equal to integral limit from 0 to 1 x upon 1 plus x square dx now we put x square equal to t so we have put x square equal to t differentiating we get 2x dx is equal to dt or we can write this x dx is equal to dt upon 2 now clearly we can see that when x is equal to lower limit that is 0 we have t is equal to 0 and when x is equal to upper limit that is 1 we have t is equal to 1 so putting these values in I1 we get I1 is equal to 1 upon 2 integral limit from 0 to 1 dt upon 1 plus t and this is equal to 1 upon 2 into now integrating 1 upon 1 plus t we get log 1 plus t and limit is from 0 to 1 and this is equal to 1 upon 2 into now putting the limits we get log 2 minus log 1 because upper limit is 1 so putting 1 in place of t we get 1 plus 1 is 2 so log 2 minus lower limit is 0 so putting 0 in place of t we get log 1 and this is equal to 1 upon 2 into log 2 because we know that log 1 is equal to 0 so we have got I1 is equal to 1 upon 2 into log 2 now we consider I2 so we have I2 is equal to integral limit from 0 to 1 x upon 1 plus x minus 1 whole square dx this is equal to integral limit from 0 to 1 x now adding and subtracting one in numerator we get x minus 1 plus 1 upon 1 plus x minus 1 whole square dx this is equal to integral limit from 0 to 1 now we can write x minus 1 plus 1 upon 1 plus x minus 1 whole square as x minus 1 upon 1 plus x minus 1 whole square plus 1 upon 1 plus x minus 1 whole square dx and this is equal to integral limit from 0 to 1 x minus 1 upon 1 plus x minus 1 whole square dx plus integral limit from 0 to 1 1 upon 1 plus x minus 1 whole square dx this is equal to now multiplying the numerator and denominator by 2 we get 1 upon 2 integral limit from 0 to 1 2 into x minus 1 upon 1 plus x minus 1 whole square dx plus integral limit from 0 to 1 1 upon 1 plus x minus 1 whole square dx and this is equal to 1 by 2 into now integral of 2 into x minus 1 upon 1 plus x minus 1 whole square is equal to log 1 plus x minus 1 whole square and limit is from 0 to 1 plus now integral of 1 upon 1 plus x minus 1 whole square is equal to tan inverse x minus 1 and limit is from 0 to 1 and this is equal to 1 by 2 into now putting the limits we get log 1 minus log 2 because we have upper limit as 1 so putting 1 in place of x here we get 0 so we get log 1 minus and the lower limit is 0 so putting here 0 we get 0 minus 1 whole square that is minus 1 whole square which is 1 so 1 plus 1 will give us 2 so we get log 2 plus here again putting the limits we get tan inverse 0 minus tan inverse minus 1 because upper limit here is 1 so putting 1 in place of x we get tan inverse 1 minus 1 that is 0 so we get tan inverse 0 minus putting lower limit 0 we get tan inverse 0 minus 1 that is tan inverse minus 1 and this is equal to 1 by 2 into now log 1 is 0 so we get 0 minus log 2 plus tan inverse 0 is 0 so we get 0 minus tan inverse minus 1 is minus pi by 4 so we get I2 is equal to minus 1 by 2 into log 2 plus pi by 4 now putting the value of I1 and I2 in I we get I is equal to pi by 4 minus 1 by 2 log 2 plus 1 by 2 log 2 minus pi by 4 and this is equal to 0 so we get I is equal to 0 thus we get our answer as B hope you have understood the solution take care and bye