IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

fuhrergurman

_{[0 ]}

^{[ 1]} tan-1 (2x-1/1+x-x²) dx

_{[0 ]}

^{[ 1]} cot-1 (1+x-x²) dx

neeraj_agarwal_1990

(1)
numerator is x+(x-1)

denominator is 1+ x(x-1)

if u observe, its the formula for tan-1x + tan-1(x-1)
now integrate it

chimanshu_007

ans -2 )
_{[0 ]}

^{[1 ]} tan^-1 (1 / 1 - x + x²)

_{[0 ]}

^{[1 ]} tan^-1 ( x + (1-x) / 1 - x + x²)

_{[0 ]}

^{[1 ]} tan^-1x + tan^-1(1-x)

_{[0 ]}

^{[1 ]} tan^-1x + _{[0 ]}

^{[1 ]}tan^-1(1-x)

= 2 _{[0 ]}

^{[1 ]}tan^-1x dx

now u can solve