#I=intsqrt(9x^2+54)dx#
#=intsqrt(9(x^2+6)dx#
#=3intsqrt(x^2+6)dx#
Let #x=sqrt6tan(theta)#
#dx=sqrt6sec(theta)^2d theta#
So:
#I=3intsqrt((sqrt6tan(theta))^2+6)*sqrt6sec(theta)^2d theta#
#=3sqrt6intsqrt(6(tan(theta)^2+1))sec(theta)^2d theta#
Because #tan(theta)^2+1=sec(theta)^2#
#I=18intsec(theta)^3d theta=18intsec(theta)*sec(theta)^2d theta#
Using Integration by parts :
#intf'(x)g(x)dx=f(x)g(x)-intf(x)g'(x)dx#
Here: #f'(x)=sec(theta)^2<=>f(x)=tan(theta)#
#g(x)=sec(theta)<=>g'(x)=tan(theta)sec(theta)#
So:
#I=18tan(theta)sec(theta)-18inttan(theta)^2sec(theta)d theta#
#=18tan(theta)sec(theta)-18int(sin(theta)^2)/(cos(theta)^3)d theta#
#=18tan(theta)sec(theta)-18int(1-cos(theta)^2)/(cos(theta)^3)d theta#
#=18tan(theta)sec(theta)-18int1/(cos(theta)^3)d theta+18int1/cos(theta)d theta#
But #I=18intsec(theta)^3d theta#
So: #I=18tan(theta)sec(theta)-I+18##intsec(theta)d theta#
#2I=18tan(theta)sec(theta)+18ln(|tan(theta)+sec(theta)|)#
#I=9tan(theta)sec(theta)+9ln(|tan(theta)+sec(theta)|)#
Finally, because #theta=tan^(-1)((sqrt6x)/6)# and that #sec(tan^(-1)(u))=sqrt(u^2+1)#,
#I=(3sqrt6)/2xsqrt(x^2/6+1)+9ln(|(sqrt6x)/6+sqrt(x^2/6+1)|)+C#, #C in RR#
\0/ Here's our answer !
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