# How do you findint x/sqrt((x^2-4x+9) )dx  using trigonometric substitution?

Apr 4, 2018

$\therefore I = 2 \ln | \frac{\left(x - 2\right) + \sqrt{{x}^{2} - 4 x + 9}}{\sqrt{5}} | + \sqrt{{x}^{2} - 4 x + 9} + C$.

#### Explanation:

Let, $I = \int \frac{x}{\sqrt{{x}^{2} - 4 x + 9}} \mathrm{dx} = \int \frac{x}{\sqrt{{\left(x - 2\right)}^{2} + {\left(\sqrt{5}\right)}^{2}}} \mathrm{dx}$.

So, we use the subst. $\left(x - 2\right) = \sqrt{5} \tan y$.

$\therefore x = 2 + \sqrt{5} \tan y , \mathmr{and} , \mathrm{dx} = \sqrt{5} {\sec}^{2} y \mathrm{dy}$.

$\therefore I = \int \frac{\left(2 + \sqrt{5} \tan y\right) \left(\sqrt{5} {\sec}^{2} y\right)}{\sqrt{5 {\tan}^{2} y + 5}} \mathrm{dy}$,

$= \int \frac{\left(2 + \sqrt{5} \tan y\right) \left(\sqrt{5} {\sec}^{2} y\right)}{\sqrt{5} \sec y} \mathrm{dy}$,

$= \int \left\{\left(2 + \sqrt{5} \tan y\right) \sec y\right\} \mathrm{dy}$,

$= 2 \int \sec y \mathrm{dy} + \sqrt{5} \int \sec y \tan y \mathrm{dy}$,

$= 2 \ln | \left(\sec y + \tan y\right) | + \sqrt{5} \sec y$.

Note that, $\left(x - 2\right) = \sqrt{5} \tan y \Rightarrow \tan y = \frac{x - 2}{\sqrt{5}} , \mathmr{and} ,$

$\sqrt{{x}^{2} - 4 x + 9} = \sqrt{5} \sec y$.

$\therefore I = 2 \ln | \frac{\left(x - 2\right) + \sqrt{{x}^{2} - 4 x + 9}}{\sqrt{5}} | + \sqrt{{x}^{2} - 4 x + 9} + C$.

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Apr 4, 2018

$I = \sqrt{{x}^{2} - 4 x + 9} + 2 \ln | \left(x - 2\right) + \sqrt{{x}^{2} - 4 x + 9} | + c$
Please see the answer given by Mr.Ratnaker Maheta for trig.substitution: $\left(x - 2\right) = \sqrt{5} \tan y .$

#### Explanation:

Here,

$I = \int \frac{x}{\sqrt{{x}^{2} - 4 x + 9}} \mathrm{dx}$

$= \frac{1}{2} \int \frac{2 x}{\sqrt{{x}^{2} - 4 x + 9}} \mathrm{dx}$

$= \frac{1}{2} \int \frac{2 x - 4 + 4}{\sqrt{{x}^{2} - 4 x + 9}} \mathrm{dx}$

$= \frac{1}{2} \int \frac{2 x - 4}{\sqrt{{x}^{2} - 4 x + 9}} \mathrm{dx} + \frac{1}{2} \int \frac{4}{\sqrt{{x}^{2} - 4 x + 9}} \mathrm{dx}$

=1/2int(d/(dx)(x^2-4x+9))/sqrt(x^2-4x+9)dx+4/2int(1)/sqrt(x^2- 4x+4+5)dx

Applying color(red)((1) given below,

$= \frac{1}{2} \cdot 2 \sqrt{{x}^{2} - 4 x + 9} + 2 \int \frac{1}{\sqrt{{\left(x - 2\right)}^{2} + {\left(\sqrt{5}\right)}^{2}}} \mathrm{dx}$

Applying color(red)((2) given below

$= \sqrt{{x}^{2} - 4 x + 9} + 2 \ln | \left(x - 2\right) + \sqrt{{\left(x - 2\right)}^{2} + {\left(\sqrt{5}\right)}^{2}} | + c$

$I = \sqrt{{x}^{2} - 4 x + 9} + 2 \ln | \left(x - 2\right) + \sqrt{{x}^{2} - 4 x + 9} | + c$

Hint:

color(red)((1)int(d/(dx)(f(x)))/sqrt(f(x))dx=2sqrt(f(x))+c

color(red)((2)int1/sqrt(X^2+A^2)dx=ln|X+sqrt(X^2+A^2)|+c