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

Aug 8, 2018

$\int \sqrt{9 {x}^{2} + 54} \mathrm{dx} = \frac{3 \sqrt{6}}{2} x \sqrt{{x}^{2} / 6 + 1} + 9 \ln \left(| \frac{\sqrt{6} x}{6} + \sqrt{{x}^{2} / 6 + 1} |\right) + C$, $C \in \mathbb{R}$

Explanation:

$I = \int \sqrt{9 {x}^{2} + 54} \mathrm{dx}$

=intsqrt(9(x^2+6)dx

$= 3 \int \sqrt{{x}^{2} + 6} \mathrm{dx}$

Let $x = \sqrt{6} \tan \left(\theta\right)$

$\mathrm{dx} = \sqrt{6} \sec {\left(\theta\right)}^{2} d \theta$

So:

$I = 3 \int \sqrt{{\left(\sqrt{6} \tan \left(\theta\right)\right)}^{2} + 6} \cdot \sqrt{6} \sec {\left(\theta\right)}^{2} d \theta$

$= 3 \sqrt{6} \int \sqrt{6 \left(\tan {\left(\theta\right)}^{2} + 1\right)} \sec {\left(\theta\right)}^{2} d \theta$

Because $\tan {\left(\theta\right)}^{2} + 1 = \sec {\left(\theta\right)}^{2}$

$I = 18 \int \sec {\left(\theta\right)}^{3} d \theta = 18 \int \sec \left(\theta\right) \cdot \sec {\left(\theta\right)}^{2} d \theta$

Using Integration by parts :

$\int f ' \left(x\right) g \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int f \left(x\right) g ' \left(x\right) \mathrm{dx}$

Here: $f ' \left(x\right) = \sec {\left(\theta\right)}^{2} \iff f \left(x\right) = \tan \left(\theta\right)$

$g \left(x\right) = \sec \left(\theta\right) \iff g ' \left(x\right) = \tan \left(\theta\right) \sec \left(\theta\right)$
So:

$I = 18 \tan \left(\theta\right) \sec \left(\theta\right) - 18 \int \tan {\left(\theta\right)}^{2} \sec \left(\theta\right) d \theta$

$= 18 \tan \left(\theta\right) \sec \left(\theta\right) - 18 \int \frac{\sin {\left(\theta\right)}^{2}}{\cos {\left(\theta\right)}^{3}} d \theta$

$= 18 \tan \left(\theta\right) \sec \left(\theta\right) - 18 \int \frac{1 - \cos {\left(\theta\right)}^{2}}{\cos {\left(\theta\right)}^{3}} d \theta$

$= 18 \tan \left(\theta\right) \sec \left(\theta\right) - 18 \int \frac{1}{\cos {\left(\theta\right)}^{3}} d \theta + 18 \int \frac{1}{\cos} \left(\theta\right) d \theta$

But $I = 18 \int \sec {\left(\theta\right)}^{3} d \theta$

So: $I = 18 \tan \left(\theta\right) \sec \left(\theta\right) - I + 18$$\int \sec \left(\theta\right) d \theta$

$2 I = 18 \tan \left(\theta\right) \sec \left(\theta\right) + 18 \ln \left(| \tan \left(\theta\right) + \sec \left(\theta\right) |\right)$

$I = 9 \tan \left(\theta\right) \sec \left(\theta\right) + 9 \ln \left(| \tan \left(\theta\right) + \sec \left(\theta\right) |\right)$

Finally, because $\theta = {\tan}^{- 1} \left(\frac{\sqrt{6} x}{6}\right)$ and that $\sec \left({\tan}^{- 1} \left(u\right)\right) = \sqrt{{u}^{2} + 1}$,

$I = \frac{3 \sqrt{6}}{2} x \sqrt{{x}^{2} / 6 + 1} + 9 \ln \left(| \frac{\sqrt{6} x}{6} + \sqrt{{x}^{2} / 6 + 1} |\right) + C$, $C \in \mathbb{R}$

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

Aug 8, 2018

$\int \left(54 + 9 {x}^{2}\right) \mathrm{dx} = \frac{3}{2} \left\{x \sqrt{6 + {x}^{2}} + 6 \ln | x + \sqrt{6 + {x}^{2}} |\right\} + C$

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)

$I = \int \sqrt{6 + {x}^{2}} \cdot 1 \mathrm{dx}$

Using Integration by parts:

$I = \sqrt{6 + {x}^{2}} \int 1 \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} \left(\sqrt{6 + {x}^{2}}\right) \int 1 \mathrm{dx}\right) \mathrm{dx}$

$I = \sqrt{6 + {x}^{2}} \times x - \int \frac{2 x}{2 \sqrt{6 + {x}^{2}}} \times x \mathrm{dx}$

$I = \sqrt{6 + {x}^{2}} \times x - \int \frac{{x}^{2}}{\sqrt{6 + {x}^{2}}} \mathrm{dx}$

$I = \sqrt{6 + {x}^{2}} \times x - \int \frac{6 + {x}^{2} - 6}{\sqrt{6 + {x}^{2}}} \mathrm{dx}$

$I = \sqrt{6 + {x}^{2}} \times x - \int \frac{6 + {x}^{2}}{\sqrt{6 + {x}^{2}}} \mathrm{dx} + \int \frac{6}{\sqrt{6 + {x}^{2}}} \mathrm{dx}$

$I = x \sqrt{6 + {x}^{2}} - \int \sqrt{6 + {x}^{2}} \mathrm{dx} + 6 \textcolor{g r e e n}{\int \frac{1}{\sqrt{{\left(\sqrt{6}\right)}^{2} + {x}^{2}}} \mathrm{dx}}$

$I = x \sqrt{6 + {x}^{2}} - \int \sqrt{6 + {x}^{2}} \mathrm{dx} + 6 \textcolor{g r e e n}{\ln | x + \sqrt{6 + {x}^{2}} |} + c$

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

$I + I = x \sqrt{6 + {x}^{2}} + 6 \ln | x + \sqrt{6 + {x}^{2}} | + c$

$\therefore 2 I = x \sqrt{6 + {x}^{2}} + 6 \ln | x + \sqrt{6 + {x}^{2}} | + c$

$\therefore I = \frac{1}{2} \left\{x \sqrt{6 + {x}^{2}} + 6 \ln | x + \sqrt{6 + {x}^{2}} |\right\} + \frac{c}{2}$

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

$\int \left(54 + 9 {x}^{2}\right) \mathrm{dx} = \frac{3}{2} \left\{x \sqrt{6 + {x}^{2}} + 6 \ln | x + \sqrt{6 + {x}^{2}} |\right\} + \frac{3 c}{2}$

$\int \left(54 + 9 {x}^{2}\right) \mathrm{dx} = \frac{3}{2} \left\{x \sqrt{6 + {x}^{2}} + 6 \ln | x + \sqrt{6 + {x}^{2}} |\right\} + C$

Note :

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