How do I evaluate the integral intsqrt(54+9x^2)dx?

2 Answers
Aug 8, 2018

intsqrt(9x^2+54)dx=(3sqrt6)/2xsqrt(x^2/6+1)+9ln(|(sqrt6x)/6+sqrt(x^2/6+1)|)+C, C in RR

Explanation:

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 !

Note : click on each blue terms to see more explanations which were secondary but still important.

Aug 8, 2018

int(54+9x^2)dx=3/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+C

Explanation:

Here ,

color(blue)(intsqrt(54+9x^2)dx=3intsqrt(6+x^2)dx=3I............(A)

Let ,

color(red)(I=intsqrt(6+x^2)dx.........................to(B)

I=intsqrt(6+x^2)*1dx

Using Integration by parts:

I=sqrt(6+x^2)int1dx-int(d/(dx)(sqrt(6+x^2))int1dx)dx

I=sqrt(6+x^2)xx x-int(2x)/(2sqrt(6+x^2)) xx xdx

I=sqrt(6+x^2)xx x-int(x^2)/(sqrt(6+x^2)) dx

I=sqrt(6+x^2)xx x-int(6+x^2-6)/(sqrt(6+x^2)) dx

I=sqrt(6+x^2)xx x-int(6+x^2)/sqrt(6+x^2)dx+int6/(sqrt(6+x^2)) dx

I=xsqrt(6+x^2)-intsqrt(6+x^2)dx+6color(green)(int1/(sqrt((sqrt6)^2+x^2)) dx)

I=xsqrt(6+x^2)-intsqrt(6+x^2)dx+6color(green)(ln|x+sqrt(6+x^2)|)+c

I=xsqrt(6+x^2)-color(red)(I)+6ln|x+sqrt(6+x^2)|+c, color(red)(....touse(B)

I+I=xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|+c

:.2I=xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|+c

:.I=1/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+c/2

From color(blue)((B) we have ,

int(54+9x^2)dx=3/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+(3c)/2

int(54+9x^2)dx=3/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+C

Note :

color(green)(int1/sqrt(X^2+a^2)dX=ln|X+sqrt(X^2+a^2)|+c