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+18intsec(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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