# How can you find the antiderivative of 1/(1+3x²) ?

Jun 11, 2018

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

#### Explanation:

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

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

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

Jun 11, 2018

$\frac{1}{\sqrt{3}} \cdot \arctan \left(\sqrt{3} \cdot x\right) + C$

#### Explanation:

Writing
$\frac{1}{1 + {\left(\sqrt{3} \cdot x\right)}^{2}}$
we Substitute
$t = \sqrt{3} x$
then we get

$\mathrm{dx} = \frac{1}{\sqrt{3}} \mathrm{dx}$
and we get

$\frac{1}{\sqrt{3}} \int \frac{1}{1 + {t}^{2}} \mathrm{dt} = \frac{1}{\sqrt{3}} \cdot \arctan \left(t\right) + C = \frac{1}{\sqrt{3}} \arctan \left(\sqrt{3} \cdot x\right) + C$

Jun 11, 2018

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

#### Explanation:

$\int \frac{1}{1 + 3 {x}^{2}} \mathrm{dx} =$ Let $x = \frac{1}{\sqrt{3}} \tan \theta$

$\tan \theta = \sqrt{3} x \therefore \theta = {\tan}^{-} 1 \sqrt{3} x$

$\therefore \mathrm{dx} = \frac{1}{\sqrt{3}} {\sec}^{2} \theta d \theta \therefore {x}^{2} = \frac{1}{3} {\tan}^{2} \theta$

$\therefore \int \frac{\mathrm{dx}}{1 + 3 {x}^{2}}$

$= \int \frac{\frac{1}{\sqrt{3}} {\sec}^{2} \theta d \theta}{1 + 3 \cdot \frac{1}{3} {\tan}^{2} \theta}$

$= \int \frac{\frac{1}{\sqrt{3}} {\sec}^{2} \theta d \theta}{1 + {\tan}^{2} \theta}$

$= \int \frac{1}{\sqrt{3}} \frac{\cancel{{\sec}^{2} \theta}}{\cancel{{\sec}^{2} \theta}} \cdot d \theta$

$= \frac{1}{\sqrt{3}} \theta + C = \frac{1}{\sqrt{3}} {\tan}^{-} 1 \sqrt{3} x + C$ ; [Ans]

Jun 11, 2018

$= - \frac{1}{\sqrt{1 + 3 {x}^{2}}} + C$

#### Explanation:

$\int \frac{1}{1 + 3 {x}^{2}} \mathrm{dx}$

$x = \frac{1}{\sqrt{3}} \tan u$

$\mathrm{dx} = \frac{1}{\sqrt{3}} \tan u \sec u \cdot \mathrm{du}$

Substitute

$\int \frac{1}{1 + 3 {x}^{2}} \mathrm{dx} = \frac{1}{\sqrt{3}} \int \frac{\sec u \tan u \cdot \mathrm{du}}{1 + {\tan}^{2} u}$

color(green)(1+tan^2u=sec^2u

=$\frac{1}{\sqrt{3}} \int \frac{\cancel{\sec u} \tan u \mathrm{du}}{\sec} ^ \cancel{2} u$

=$\frac{1}{\sqrt{3}} \int \sin u \cdot \mathrm{du}$

=$- \frac{1}{\sqrt{3}} \cos u + C$

reverse the substitution

$= - \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{1 + 3 {x}^{2}}} + C$