# How do you integrate int (x^2-1)/sqrt(x^2+9)dx using trigonometric substitution?

Aug 5, 2018

$\int \frac{{x}^{2} - 1}{\sqrt{{x}^{2} + 9}} \mathrm{dx} = \frac{x}{2} \sqrt{{x}^{2} + 9} - \frac{11}{2} \ln | x + \sqrt{{x}^{2} + 9} | + C$

#### Explanation:

Here ,

$\int \frac{{x}^{2} - 1}{\sqrt{{x}^{2} + 9}} \mathrm{dx} = \int \frac{{x}^{2} + 9 - 10}{\sqrt{{x}^{2} + 9}} \mathrm{dx}$

$\int \frac{{x}^{2} - 1}{\sqrt{{x}^{2} + 9}} \mathrm{dx} = \int \sqrt{{x}^{2} + 9} \mathrm{dx} - 10 \int \frac{1}{\sqrt{{x}^{2} + 9}} \mathrm{dx}$

int(x^2-1)/sqrt(x^2+9)dx=I-10ln|x+sqrt(x^2+9)|color(red)(...to(A)

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

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

Using Integration by parts:

$I = \sqrt{{x}^{2} + 9} \int 1 \mathrm{dx} - \int \left(\frac{1}{2 \sqrt{{x}^{2} + 9}} \left(2 x\right) \int 1 \mathrm{dx}\right) \mathrm{dx}$

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

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

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

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

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

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

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

From eqn color(red)((A) we have

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

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 10 \ln | x + \sqrt{{x}^{2} + 9} | + \frac{c}{2}$

$\int \frac{{x}^{2} - 1}{\sqrt{{x}^{2} + 9}} \mathrm{dx} = \frac{x}{2} \sqrt{{x}^{2} + 9} - \frac{11}{2} \ln | x + \sqrt{{x}^{2} + 9} | + C$