AP EAMCET202124 Aug 2021Morning ShiftMathematicsIndefinite IntegrationActual
The value of e^ ⁻¹(x) 1+x^2 [ ( ⁻¹ 1+x^2 )^2+ ⁻¹ ( 1-x^2 1+x^2 ) ] d x , for x>0 is
Options
- Ae^ ⁻¹(x) ( ⁻¹ x )^2+c
- Be^ ⁻¹(x) ( ⁻¹ x )+c
- Ce^ ⁻¹(x) ( ⁻¹ x )^3+c
- D-e^ ⁻¹(x) ( ⁻¹ x )^2+c
Correct answer
A. e^ ⁻¹(x) ( ⁻¹ x )^2+c
Step-by-step solution
Let I= e^ ⁻¹ x 1+x^2 [ ( ⁻¹ 1+x^2 )^2+ ⁻¹ ( 1-x^2 1+x^2 ) ] d x(x>0) Let us take ⁻¹ x= x= aligned & 1 1+x^2 d x=d & I = e^ [ ( ⁻¹ ( 1+ ^2 ) )^2 . & .+ ⁻¹ ( 1- ^2 1+ ^2 ) ] d & = e^ ( ⁻¹ )^2+ ( ⁻¹ 2 ) d & [ 2 = 1- ^2 1+ ^2 ] & = e^ ^2+2 d [ ⁻¹ = . and & aligned aligned & ⁻¹ 2 =2 ] & [ e^x (f(x)+f^ (x) ] d x=e^x f(x)+c ] aligned So, aligned & I=e^ ^2+c & I=e^ ⁻¹ x ( ⁻¹ x )^2+c aligned