Question

2.V1+4x2

Answer 1

The given function is 1 1+4 x 2 .

Consider, ∫ 1 1+4 x 2 dx(1)

Also, ∫ 1 x 2 + a 2 =log| x+ x 2 + a 2 | +c (2)

Now let, 2 x =t 2dx=dt

Substitute values of t and dt in (1),

∫ 1 1+4 x 2 dx= 1 2 ∫ dt 1+ t 2 = 1 2 log| t+ 1+ t 2 |+c by (2)

Now substitute t as 2x in (2),

∫ 1 1+4 x 2 dx= 1 2 log| 2x+ 4 x 2 +1 |+c

Thus, the integral of the function 1 1+4 x 2 is 1 2 log| 2x+ 4 x 2 +1 |+c.