# How do i integrate x^4/(x^2+1)^3 ?

Apr 11, 2018

$I = \frac{1}{8} \left[3 {\tan}^{-} 1 x - \frac{5 {x}^{3} + 3 x}{{\left(1 + {x}^{2}\right)}^{2}}\right]$

#### Explanation:

Here,

$I = \int {x}^{4} / {\left({x}^{2} + 1\right)}^{3} \mathrm{dx}$

Let, $x = \tan t \implies \mathrm{dx} = {\sec}^{2} t \mathrm{dt}$

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

So,$I = \int {\tan}^{4} \frac{t}{{\sec}^{2} t} ^ 3 {\sec}^{2} t \mathrm{dt}$

=inttan^4t/sec^4tdt=int(sin^4t/cos^4t)/(1/cos^4t)dt=intsin^4tdt

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

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

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

$= \frac{1}{8} \int \left(3 - 4 \cos 2 t + \cos 4 t\right) \mathrm{dt}$

$= \frac{1}{8} \left[3 t - \frac{4 \sin 2 t}{2} + \frac{\sin 4 t}{4}\right] + c$

$= \frac{1}{8} \left[3 t - 2 \left(\frac{2 \tan t}{1 + {\tan}^{2} t}\right) + \frac{1}{4} \left(2 \sin 2 t \cos 2 t\right)\right] + c$

=1/8[3t-(4tant)/(1+tan^2t)+1/2((2tant)/(1+tan^2t))((1- tan^2t)/(1+tan^2t))]

Substituting back $x = \tan t \implies t = {\tan}^{-} 1 x$,we get

$= \frac{1}{8} \left[3 {\tan}^{-} 1 x - \frac{4 x}{1 + {x}^{2}} + \frac{x}{1 + {x}^{2}} \times \frac{1 - {x}^{2}}{1 + {x}^{2}}\right] + c$

$I = \frac{1}{8} \left[3 {\tan}^{-} 1 x - \left(\frac{4 x \left(1 + {x}^{2}\right) - x \left(1 - {x}^{2}\right)}{1 + {x}^{2}} ^ 2\right)\right]$

$I = \frac{1}{8} \left[3 {\tan}^{-} 1 x - \frac{5 {x}^{3} + 3 x}{{\left(1 + {x}^{2}\right)}^{2}}\right]$

Apr 11, 2018

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3 \arctan x}{8} - \frac{5 {x}^{3} + 3 x}{8 {\left(1 + {x}^{2}\right)}^{2}} + C$

#### Explanation:

Let:

$x = \tan t$

with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$\mathrm{dx} = {\sec}^{2} t \mathrm{dt}$

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

Then:

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \int \frac{{\tan}^{4} t {\sec}^{2} t \mathrm{dt}}{\sec} ^ 6 t$

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

and as:

$\tan \frac{t}{\sec} t = \frac{\sin \frac{t}{\cos} t}{\frac{1}{\cos} t} = \sin t$

then:

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

Write the integrand as:

${\sin}^{4} t = {\sin}^{2} t \left(1 - {\cos}^{2} t\right) = {\sin}^{2} t - {\sin}^{2} t {\cos}^{2} t$

${\sin}^{4} t = \frac{1}{2} - \frac{\cos 2 t}{2} - \frac{{\sin}^{2} 2 t}{4}$

${\sin}^{4} t = \frac{1}{2} - \frac{\cos 2 t}{2} - \frac{1 - \cos 4 t}{8}$

${\sin}^{4} t = \frac{3}{8} - \frac{\cos 2 t}{2} + \frac{\cos 4 t}{8}$

Then using the linearity of the integral:

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3}{8} \int \mathrm{dt} - \frac{1}{2} \int \cos 2 t \mathrm{dt} + \frac{1}{8} \int \cos 4 t \mathrm{dt}$

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3 t}{8} - \frac{\sin 2 t}{4} + \frac{\sin 4 t}{32} + C$

To undo the substitution use the trigonometric formulas:

$\sin 4 t = 2 \sin 2 t \cos 2 t$

to have:

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3 t}{8} - \frac{\sin 2 t}{4} + \frac{\sin 2 t \cos 2 t}{16} + C$

and then the parametric formulas:

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

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

so:

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3 \arctan x}{8} - \frac{x}{2 \left(1 + {x}^{2}\right)} + \frac{x \left(1 - {x}^{2}\right)}{8 {\left(1 + {x}^{2}\right)}^{2}} + C$

and simplifying:

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3 \arctan x}{8} - \frac{4 x \left(1 + {x}^{2}\right) - x \left(1 - {x}^{2}\right)}{8 {\left(1 + {x}^{2}\right)}^{2}} + C$

$\int \frac{{x}^{4} \mathrm{dx}}{1 + {x}^{2}} ^ 3 = \frac{3 \arctan x}{8} - \frac{5 {x}^{3} + 3 x}{8 {\left(1 + {x}^{2}\right)}^{2}} + C$