How do you integrate #sqrt(4x² + 1)#?

3 Answers
Aug 14, 2018

#:.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#

Aug 14, 2018

#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#

Aug 14, 2018

#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#