# How do you integrate int 1/sqrt(4x^2-12x+34)  using trigonometric substitution?

Jun 14, 2018

$= \frac{1}{2} \ln | \sqrt{4 {x}^{2} - 12 x + 34} + \left(2 x - 3\right) | + C$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 34}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 9 + 25}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} + {5}^{2}}} \mathrm{dx}$

Subst, $2 x - 3 = 5 \tan u \implies 2 \mathrm{dx} = 5 {\sec}^{2} u \mathrm{du}$

=>dx=5/2sec^2udu and color(blue)(tanu=(2x-3)/5

So

$I = \int \frac{1}{\sqrt{{5}^{2} {\tan}^{2} u + {5}^{2}}} \times \frac{5}{2} {\sec}^{2} u \mathrm{du}$

$= \frac{5}{2} \int {\sec}^{2} \frac{u}{5 \sec u} \mathrm{du}$

$= \frac{1}{2} \int \sec u \mathrm{du}$

$= \frac{1}{2} \ln | \sec u + \tan u | + c$

$= \frac{1}{2} \ln | \sqrt{{\tan}^{2} u + 1} + \tan u | + c$

Subst. back , color(blue)(tanu=(2x-3)/5 we get

$I = \frac{1}{2} \ln | \sqrt{{\left(\frac{2 x - 3}{5}\right)}^{2} + 1} + \left(\frac{2 x - 3}{5}\right) | + c$

$= \frac{1}{2} \ln | \frac{\sqrt{4 {x}^{2} - 12 x + 34}}{5} + \frac{2 x - 3}{5} | + c$

$= \frac{1}{2} \ln | \frac{\left(\sqrt{4 {x}^{2} - 12 x + 34}\right) + \left(2 x - 3\right)}{5} | + c$

$= \frac{1}{2} \ln | \sqrt{4 {x}^{2} - 12 x + 34} + \left(2 x - 3\right) | - \frac{1}{2} \ln 5 + c$

$= \frac{1}{2} \ln | \sqrt{4 {x}^{2} - 12 x + 34} + \left(2 x - 3\right) | + C$

$w h e r e , C = - \frac{1}{2} \ln 5 + c$

Jun 14, 2018

Integration without trigonometric substitution.
$I = \frac{1}{2} \ln | \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x + 34} | + c$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 34}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 9 + 25}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} + {5}^{2}}} \mathrm{dx}$

Subst, $\textcolor{b l u e}{2 x - 3 = u} \implies 2 \mathrm{dx} = \mathrm{du} \implies \mathrm{dx} = \frac{1}{2} \mathrm{du}$

So,

$I = \int \frac{1}{\sqrt{{u}^{2} + {5}^{2}}} \times \frac{1}{2} \mathrm{du}$

=1/2color(red)(int1/sqrt(u^2+5^2)du

=1/2color(red)(ln|u+sqrt(u^2+5^2)|+c

Subst. back ,color(blue)(u=2x-3  ,we get

$I = \frac{1}{2} \ln | \left(2 x - 3\right) + \sqrt{{\left(2 x - 3\right)}^{2} + {5}^{2}} | + c$

$= \frac{1}{2} \ln | \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x + 34} | + c$
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Note :

color(red)((1)int1/sqrt(x^2+a^2)dx=ln|x+sqrt(x^2+a^2)|+c