# How do you find the integral of sqrt(x²-49)dx?

Apr 22, 2018

$I = \sqrt{{x}^{2} - {7}^{2}} \mathrm{dx}$
We know that,
color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-a^2/2ln|x+sqrt(x^2-a^2)|+c,
put $x = 7$
$I = \frac{x}{2} \sqrt{{x}^{2} - 49} - \frac{49}{2} \ln | x + \sqrt{{x}^{2} - 49} | + c$

#### Explanation:

We can solve above question $\text{without using"color(red)" Result(1).}$

$I = \int \sqrt{{x}^{2} - 49} \mathrm{dx} \ldots \to \left(A\right)$

$= \int 1 \cdot \sqrt{{x}^{2} - 49} \mathrm{dx}$

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

$\int \left(u \cdot v\right) \mathrm{dx} = u \int v \mathrm{dx} - \int \left(u ' \int v \mathrm{dx}\right) \mathrm{dx}$

$u = \sqrt{{x}^{2} - 49} \mathmr{and} v = 1$

$\implies u ' = \frac{1}{2 \sqrt{{x}^{2} - 49}} 2 x = \frac{x}{\sqrt{{x}^{2} - 49}} \mathmr{and} \int v \mathrm{dx} = x$

So,

$I = \sqrt{{x}^{2} - 49} \cdot x - \int \frac{x}{\sqrt{{x}^{2} - 49}} \cdot x \mathrm{dx}$

$= x \cdot \sqrt{{x}^{2} - 49} - \int {x}^{2} / \sqrt{{x}^{2} - 49} \mathrm{dx}$

$= x \sqrt{{x}^{2} - 49} - \int \frac{\left({x}^{2} - 49\right) + 49}{\sqrt{{x}^{2} - 49}}$

$I = x \sqrt{{x}^{2} - 49} - \int \sqrt{{x}^{2} - 49} \mathrm{dx} - \int \frac{49}{\sqrt{{x}^{2} - 49}} \mathrm{dx}$

$I = x \sqrt{{x}^{2} - 49} - I - 49 \int \frac{1}{\sqrt{{x}^{2} - {7}^{2}}} \mathrm{dx} \ldots \to F r o m \left(A\right)$

$I + I = x \sqrt{{x}^{2} - 49} - 49 \ln | x + \sqrt{{x}^{2} - {7}^{2}} | + c$

$2 I = x \sqrt{{x}^{2} - 49} - 49 \ln | x + \sqrt{{x}^{2} - 49} | + c$

$\implies I = \frac{x}{2} \sqrt{{x}^{2} - 49} - \frac{49}{2} \ln | x + \sqrt{{x}^{2} - 49} | + c$