Question

Integrate the following with respect to x:

`tan^-1 ((8x)/(1 - 16x^2))`

Answer 1

`int tan^-1 ((8x)/(1 - 16x^2))  "d"x = int tan^-1 ((2 xx 4x)/(1 - (4x)^2)) "d"x`

Put 4x = tan θ

4 dx = sec²θ dθ

`int tan^-1 ((8x)/(1 - 16x^2))  "d"x = inttan^-1  ((2 tan theta)/(1 - tan^2theta)) (sec^2theta)/4  "d"theta`

= `1/4 int tan^-1 (tan 2theta) sec^2 theta  "d"theta`

= `1/4 int 2theta sin^2theta  "d"theta`

= `1/2 int theta sec^2theta  "d"theta`  .........(1)

Consider `int theta sec^2theta  "d"theta`

u = θ

u' = 1

u" = 0

dv = `sin^2theta  "d"theta`

⇒ v = `int tan theta  "d"theta`

⇒ v = tan θ

v₁ = `int "v"  "d"theta`

= `int tan theta  "d"theta`

= `log |sec theta|`

v₂ = `int "v"_1  "d"theta`

= `int log |sec theta|  "d"theta`

`int "u"  "dv"` = uv – u'v₁ + u"v₂ – .............

`int theta sec^2theta  "d"theta = theta tan theta - 1 xx log |sec theta| + 0 xx int log |sec theta|  "d"theta + "c"`

`int theta sec^2theta  "d"theta = theta tan theta - log |sec theta| + "c"`

Substituting in eqquation (1) we get

`int tan^-1 ((8x)/(1 - 16x^2))  "d"x = 1/2 [theta tan theta - log |sec theta|] + "c"`   ........(2)

4x = tan θ

⇒ θ = `tan^-1 (4x)`

sec θ = `sqrt(1 + tan^2theta)`

= `sqrt(1 + (4x)^2`

= `sqrt(1 + 16x^2)`

Substituting in equation (2) we get

`int tan^-1 ((8x)/(1 - 16x^2))  "d"x = 1/2 [tan^-1 (4x) xx 4x - log |sqrt(1 + 16x^2)|] + "c"`

= `1/2 [4x tan^-1 (4x) - log |sqrt(1 + 16x^2)|] + "c"`