If ^3 x ( ⁻¹( x+ x) )⁻¹ ( ^4 x+3 ^2 x+1 ) d x=f(x)+C , then e^ f (x) =

Step-by-step solution

I= ^3(x) ( ^4 x+3 ^2 x+1 ) ( ⁻¹ cosec x+ x ) d x Let ⁻¹( x+ x)=t aligned & 1 1+( x+ x)^2 [ x x- x] d x=d t & ^2 x ^2 x+ ^4 x+2 ^2 x+1 & [ x- x ^2 x ^2 x ] d x=d t & = x (1- ^2 x ) ^2 x+ ^4 x+2 ^2 x+1 d x=d x & = ^3 x ^4 x+3 ^2 x+1 d x=d t & = d t t & = (t)+c & = | ⁻¹ csec ^2(x)+ (x) |+c & e^ f(x) = ⁻¹( x+ x)+c aligned