Question

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

Answer 1

`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+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 !

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

Answer 2

`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`