# How do you integrate int dx/(x^2+1)^2 using trig substitutions?

Aug 4, 2016

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

#### Explanation:

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

Use $x = \tan \left(a\right)$

$\mathrm{dx} = {\sec}^{2} \left(a\right) \mathrm{da}$

$\int \frac{\mathrm{dx}}{{x}^{2} + 1} ^ 2 = \int \frac{{\sec}^{2} \left(a\right) \mathrm{da}}{1 + {\tan}^{2} a} ^ 2$

Use the identity $1 + {\tan}^{2} \left(a\right) = {\sec}^{2} \left(a\right)$

$\int \frac{\mathrm{dx}}{{x}^{2} + 1} ^ 2 = \int \frac{{\sec}^{2} \left(a\right) \mathrm{da}}{\sec} ^ 4 \left(a\right)$

$= \int \frac{\mathrm{da}}{\sec} ^ 2 \left(a\right)$

$= \int {\cos}^{2} \left(a\right) \mathrm{da}$
$= \int \left(\frac{1 + \cos \left(2 a\right)}{2}\right) \mathrm{da}$

$= \left(\frac{1}{2}\right) \left(\int \left(\mathrm{da}\right) + \int \cos \left(2 a\right) \mathrm{da}\right)$
$= \left(\frac{1}{2}\right) \left(a + \sin \frac{2 a}{2}\right)$
$= \left(\frac{1}{2}\right) \left(a + \frac{2 \sin \left(a\right) \cos \left(a\right)}{2}\right)$
$= \left(\frac{1}{2}\right) \left(a + \sin \left(a\right) . \cos \left(a\right)\right)$

we know that $a = {\tan}^{-} 1 \left(x\right)$

sin(a)=x/(sqrt(1+x^2)
cos(a)=x/(sqrt(1+x^2

int dx/(x^2+1)^2= (1/2) (tan^-1(x) +sin(sin^-1(x/(sqrt(1+x^2)))cos(cos^-1(1/(sqrt(1+x^2))))
=(1/2) (tan^-1(x)+(x/(sqrt(1+x^2))1/sqrt(1+x^2))

$= \left(\frac{1}{2}\right) \left({\tan}^{-} 1 \left(x\right) + \frac{x}{1 + {x}^{2}}\right)$

Aug 4, 2016

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

#### Explanation:

$\int \frac{\mathrm{dx}}{{x}^{2} + 1} ^ 2$ performing the substitution

$x = \tan \left(y\right)$ and consequently
$\mathrm{dx} = \frac{\mathrm{dy}}{\cos {\left(y\right)}^{2}}$

we have

$\int \frac{\mathrm{dx}}{{x}^{2} + 1} ^ 2 \equiv \int \frac{\mathrm{dy}}{\cos {\left(y\right)}^{2} \left(\frac{1}{\cos} {\left(y\right)}^{4}\right)} = \int \cos {\left(y\right)}^{2} \mathrm{dy}$

but

$\frac{d}{\mathrm{dy}} \left(\sin \left(y\right) \cos \left(y\right)\right) = \cos {\left(y\right)}^{2} - \sin {\left(y\right)}^{2} = 2 \cos {\left(y\right)}^{2} - 1$

then

$\int \cos {\left(y\right)}^{2} \mathrm{dy} = \frac{1}{2} \left(y + \sin \left(y\right) \cos \left(y\right)\right)$

Finally, recalling $y = \arctan \left(x\right)$ we have

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