How do you find the integral of #sqrt(x^2+9) dx#?

2 Answers
Jun 2, 2018

#-1/2*(t^2/2+18ln(t)-81/82t^3)+C# where #t=sqrt(x^2+9)-x#

Explanation:

Setting
#sqrt(x^2+9)=t+x#
then we get
#x=(9-t^2)/(2*t)#
and
#dx=-(t^2+9)/(2t^2)dt#
so we get#-1/2int (t^2+9)^2/t^3dt#
this is
#-1/2int (t+18/t+81/t^3)dt=#
#-1/2*(t^2/2+18ln(t)-81/(2t^2))+C#

Jun 2, 2018

We know that,

#intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c#

Taking, #a=3# , we get

#intsqrt(x^2+9)dx=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+c#

Explanation:

#II^(nd) Method#

Here,

#I=sqrt(x^2+9) dx...to(1)#

#I=intsqrt(x^2+9)*1dx#

#"Using "color(blue)"Integration by Parts :"#

#color(blue)(intu*vdx=uintvdx-int(u'intvdx)dx#

Let #u=sqrt(x^2+9) and v=1#

#=>u'=(2x)/(2sqrt(x^2+9))=x/sqrt(x^2+9) and intvdx=x+c#

So,

#I=sqrt(x^2+9)*x-intx/sqrt(x^2+9)*xdx#

#I=xsqrt(x^2+9)-intx^2/sqrt(x^2+9)dx#

#I=xsqrt(x^2+9)-int((x^2+9)-9)/sqrt(x^2+9)dx#

#I=xsqrt(x^2+9)-int(x^2+9)/sqrt(x^2+9)dx+int9/sqrt(x^2+9)dx#

#I=xsqrt(x^2+9)-intsqrt(x^2+9)dx+9int1/sqrt(x^2+3^2)dx#

#I=xsqrt(x^2+9)-I+9ln|x+sqrt(x^2+3^2)|+Cto#from #(1)#

#I+I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+C#

#2I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+C#

#I=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+C#