₀^ / 2 d x 5+4 x =A ⁻¹ B , then A+B=

Step-by-step solution

Let I= ₀^ / 2 d x 5+4 x Put x 2 = t ^2 x 2 ( 1 2 ) dx = dt dx = 2 dt 1+ t ^2 and x = 2 t 1+ t ^2 When x =0, t =0 and when x = 2 , t =1 aligned & = ₀^1 1 5+4 ( 2 t 1+ t ^2 ) 2 dt 1+ t ^2 =2 ₀^1 dt 5+5 t ^2+8 t = 2 5 ₀^1 dt t ^2+ 8 5 t +1 & = 2 5 ₀^1 dt t ^2+ 8 5 t + 16 25 + 9 25 = 2 5 ₀^1 dt ( t + 4 5 )^2+ ( 3 5 )^2 aligned aligned & I= 2 5 1 ( 3 5 ) [ ⁻¹ [ t + 4 5 ( 3 5 ) ] ]₀^1= 2 3 [ ⁻¹ ( 5 t +4 3 ) ]₀^1 & = 2 3 [ ⁻¹ 3- ⁻¹ 4 3 ]= 2 3 ⁻¹ [ 3- ( 4 3 ) 1+3 ( 4 3 ) ] & = 2 3 ⁻¹ ( 5 3 1 5 )= 2 3 ⁻¹ ( 1 3 ) aligned Com