Question

How do you integrate `sqrt(4x² + 1)`?

Answer 1

`:.I=x/2sqrt(4x^2+1)+1/4ln|2x+sqrt(4x^2+1)|+c`

Explanation:

We know that,

`color(red)((1)intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c`

Let.

`I=intsqrt(4x^2+1)dx`

Substitute `2x=u=>x=u/2=>dx=1/2du`

`I=intsqrt(u^2+1)*1/2du`

`:.I=1/2intsqrt(u^2+1^2)du`

Using `(1)` we get

`I=1/2{u/2sqrt(u^2+1^2)+1^2/2ln|u+sqrt(u^2+1^2)|}+c`

Subst. back `u=2x`

`I=1/2{(2x)/2sqrt(4x^2+1)+1/2ln|2x+sqrt(4x^2+1)|}+c`

`:.I=x/2sqrt(4x^2+1)+1/4ln|2x+sqrt(4x^2+1)|+c`

Answer 2

`I=x/2sqrt(4x^2+1)+1/4ln|2x+sqrt(4x^2+1)|+C`

Explanation:

Here,

`I=intsqrt(4x^2+1)dx................to(1)`

`I=intsqrt(4x^2+1)*1dx`

Using Integration by parts:

`I=sqrt(4x^2+1)int1dx-int(d/(dx)(sqrt(4x^2+1))int1dx)dx`

`:.I=sqrt(4x^2+1)*x-int(8x)/(2sqrt(4x^2+1)) xx xdx`

`:.I=xsqrt(4x^2+1)-int(4x^2)/sqrt(4x^2+1) dx`

`:.I=xsqrt(4x^2+1)-int(4x^2+1-1)/sqrt(4x^2+1) dx`

`:.I=xsqrt(4x^2+1)-intsqrt(4x^2+1) dx+int1/sqrt(4x^2+1)dx`

`:.I=xsqrt(4x^2+1)-I+int1/sqrt((2x)^2+1)to[because (1)]`

`:.I+I=xsqrt(4x^2+1)+1/2ln|2x+sqrt((2x)^2+1)|+c`

`:.2I=xsqrt(4x^2+1)+1/2ln|2x+sqrt(4x^2+1)|+c`

`:.I=x/2sqrt(4x^2+1)+1/4ln|2x+sqrt(4x^2+1)|+C`

Answer 3

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

Explanation:

One method is to use a hyperbolic substitution with `x=1/2sinh u`

Then:

`int \ sqrt(4x^2+1) \ dx = int \ sqrt(sinh^2 u + 1) \ (dx)/(du) \ du`

`color(white)(int \ sqrt(4x^2+1) \ dx) = int \ sqrt(sinh^2 u + 1) 1/2 cosh u \ du`

`color(white)(int \ sqrt(4x^2+1) \ dx) = int \ sqrt(cosh^2 u) 1/2 cosh u \ du`

`color(white)(int \ sqrt(4x^2+1) \ dx) = int \ 1/2 cosh^2 u \ du`

`color(white)(int \ sqrt(4x^2+1) \ dx) = int \ 1/4 (1 + cosh 2 u) \ du`

`color(white)(int \ sqrt(4x^2+1) \ dx) = 1/4u + 1/8 sinh 2u + C`

`color(white)(int \ sqrt(4x^2+1) \ dx) = 1/4u + 1/4 sinh u cosh u + C`

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