# How do you use substitution to integrate (4x^3)/( (x^2+1)^2)?

Mar 5, 2018

I tried this:

#### Explanation:

Have a look:

Mar 5, 2018

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = 2 \ln \left(1 + {x}^{2}\right) + \frac{2}{1 + {x}^{2}} + C$

#### Explanation:

Substitute $x = \tan t$, $\mathrm{dx} = {\sec}^{2} t$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{4 {\tan}^{3} t {\sec}^{2} t}{{\tan}^{2} t + 1} ^ 2 \mathrm{dt}$

Use now the trigonometric identity:

$1 + {\tan}^{2} t = {\sec}^{2} t$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{4 {\tan}^{3} t {\sec}^{2} t}{\sec} ^ 4 t \mathrm{dt}$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{4 {\tan}^{3} t}{\sec} ^ 2 t \mathrm{dt}$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{4 {\sin}^{3} t {\cos}^{2}}{\cos} ^ 3 t \mathrm{dt}$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{4 {\sin}^{3} t}{\cos} t \mathrm{dt}$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = 4 \int \frac{\left(1 - {\cos}^{2} t\right) \sin t}{\cos} t \mathrm{dt}$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = - 4 \int \left(\frac{1}{\cos} t - \cos t\right) d \left(\cos t\right)$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = - 4 \ln \left\mid \cos t \right\mid + 2 {\cos}^{2} t + C$

using the properties of logarithms:

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = 2 \ln \left(\frac{1}{\cos} ^ 2 t\right) + 2 {\cos}^{2} t + C$

To undo the substitution note that:

${\cos}^{2} t = \frac{1}{\sec} ^ 2 t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + {x}^{2}}$

$\int \frac{4 {x}^{3}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = 2 \ln \left(1 + {x}^{2}\right) + \frac{2}{1 + {x}^{2}} + C$