How do you find the integral of sqrt(x^2+9) dx?

Jun 2, 2018

$- \frac{1}{2} \cdot \left({t}^{2} / 2 + 18 \ln \left(t\right) - \frac{81}{82} {t}^{3}\right) + C$ where $t = \sqrt{{x}^{2} + 9} - x$

Explanation:

Setting
$\sqrt{{x}^{2} + 9} = t + x$
then we get
$x = \frac{9 - {t}^{2}}{2 \cdot t}$
and
$\mathrm{dx} = - \frac{{t}^{2} + 9}{2 {t}^{2}} \mathrm{dt}$
so we get$- \frac{1}{2} \int {\left({t}^{2} + 9\right)}^{2} / {t}^{3} \mathrm{dt}$
this is
$- \frac{1}{2} \int \left(t + \frac{18}{t} + \frac{81}{t} ^ 3\right) \mathrm{dt} =$
$- \frac{1}{2} \cdot \left({t}^{2} / 2 + 18 \ln \left(t\right) - \frac{81}{2 {t}^{2}}\right) + C$

Jun 2, 2018

We know that,

$\int \sqrt{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{x}{2} \sqrt{{x}^{2} + {a}^{2}} + {a}^{2} / 2 \ln | x + \sqrt{{x}^{2} + {a}^{2}} | + c$

Taking, $a = 3$ , we get

$\int \sqrt{{x}^{2} + 9} \mathrm{dx} = \frac{x}{2} \sqrt{{x}^{2} + 9} + \frac{9}{2} \ln | x + \sqrt{{x}^{2} + 9} | + c$

Explanation:

$I {I}^{n d} M e t h o d$

Here,

$I = \sqrt{{x}^{2} + 9} \mathrm{dx} \ldots \to \left(1\right)$

$I = \int \sqrt{{x}^{2} + 9} \cdot 1 \mathrm{dx}$

$\text{Using "color(blue)"Integration by Parts :}$

color(blue)(intu*vdx=uintvdx-int(u'intvdx)dx

Let $u = \sqrt{{x}^{2} + 9} \mathmr{and} v = 1$

$\implies u ' = \frac{2 x}{2 \sqrt{{x}^{2} + 9}} = \frac{x}{\sqrt{{x}^{2} + 9}} \mathmr{and} \int v \mathrm{dx} = x + c$

So,

$I = \sqrt{{x}^{2} + 9} \cdot x - \int \frac{x}{\sqrt{{x}^{2} + 9}} \cdot x \mathrm{dx}$

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

$I = x \sqrt{{x}^{2} + 9} - \int \frac{\left({x}^{2} + 9\right) - 9}{\sqrt{{x}^{2} + 9}} \mathrm{dx}$

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

$I = x \sqrt{{x}^{2} + 9} - \int \sqrt{{x}^{2} + 9} \mathrm{dx} + 9 \int \frac{1}{\sqrt{{x}^{2} + {3}^{2}}} \mathrm{dx}$

$I = x \sqrt{{x}^{2} + 9} - I + 9 \ln | x + \sqrt{{x}^{2} + {3}^{2}} | + C \to$from $\left(1\right)$

$I + I = x \sqrt{{x}^{2} + 9} + 9 \ln | x + \sqrt{{x}^{2} + 9} | + C$

$2 I = x \sqrt{{x}^{2} + 9} + 9 \ln | x + \sqrt{{x}^{2} + 9} | + C$

$I = \frac{x}{2} \sqrt{{x}^{2} + 9} + \frac{9}{2} \ln | x + \sqrt{{x}^{2} + 9} | + C$