# What is the integral of sqrt(1+4x^2) ?

Find `int sqrt(1+4x^2)dx` :

(1) Let `x=(tanu)/2` . Then `dx=1/2 sec^2udu` and `u=tan^(-1)(2x)`

Now substituting for x we get:

`int sqrt(1+4x^2)dx=int sqrt(1+tan^2u)1/2sec^2u du=1/2int sec^3udu`

(2) To evaluate `1/2 int sec^3u du` we use integration by parts:

`int u dv=uv-int v du`

Let `u=secx` , then `du=secxtanxdx`

Let `v=tanx` , then `dv=sec^2xdx`

Thus `int sec^3udu=secutanu-int secutan^2udu`

`=secutanu-int secu(sec^2u-1)du`

`=secutanu-intsec^3udu+int secudu`

`2intsec^3udu=secutanu+int secudu`

`int sec^3udu=1/2[secutanu+ln|secu+tanu|+C`

(3) Now `1/2 int sec^3udu=1/4[secutanu+ln|secu+tanu|]+C`

Substituting back for x we get:

` `** Note that `secutanu=sqrt(tan^2u+1)tanu` and `tanu=2x` **

`1/4[secutanu+ln|secu+tanu|]+C`

`=1/4[2xsqrt(4x^2+1)+ln|sqrt(4x^2+1)+2x|]+C`

** Note that `sinh^(-1)(2x)=ln(2x+sqrt(4x^2+1))` **

`=1/2xsqrt(4x^2+1)+1/4 sinh^(-1)(2x)+C`

(4) Thus the solution is:

`int sqrt(1+4x^2)dx=1/2xsqrt(4x^2+1)+1/4sinh^(-1)(2x)+C`

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