# How do you integrate int (3x) / (x^2 * (x^2+1) ) using partial fractions?

Apr 5, 2017

The answer is $= 3 \ln \left(| x |\right) - \frac{3}{2} \ln \left({x}^{2} + 1\right) + C$

#### Explanation:

The expression needs some simplification

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

Now we can perform the decomposition into partial fractions

$\frac{1}{x \left({x}^{2} + 1\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 1}$

$= \frac{A \left({x}^{2} + 1\right) + x \left(B x + C\right)}{x \left({x}^{2} + 1\right)}$

The denominators are the same, we compare the numerators

$1 = \left(A \left({x}^{2} + 1\right) + x \left(B x + C\right)\right)$

Let $x = 0$, $\implies$, $1 = A$

Coefficients of ${x}^{2}$,

$0 = A + B$

$B = - 1$

Coefficients of $x$,

$0 = C$

Therefore,

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

So,

the integral is

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

The second integral is done by substutution

Let $u = {x}^{2} + 1$

$\mathrm{du} = 2 x \mathrm{dx}$

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

$= \frac{1}{2} \ln u$

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

Putting it all together,

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

Apr 5, 2017

$\frac{3}{2} \ln \left\{{x}^{2} / \left({x}^{2} + 1\right)\right\} + C , \mathmr{and} ,$

$3 \ln | x | - \frac{3}{2} \ln \left({x}^{2} + 1\right) + C .$

#### Explanation:

Resprcted Narad T. has solved the Problem using Partial

Fractions, as is reqd. to be done in that fashion.

As an Aliter, I submit the following Soln. :

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

Noting that, $\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 x ,$ we subst. ${x}^{2} = t , \text{ so, } 2 x \mathrm{dx} = \mathrm{dt} .$

$\therefore I = \frac{3}{2} \int \frac{1}{t \left(t + 1\right)} \mathrm{dt} = \frac{3}{2} \int \frac{\left(t + 1\right) - \left(t\right)}{t \left(t + 1\right)} \mathrm{dt} ,$

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

$= \frac{3}{2} \int \left\{\frac{1}{t} - \frac{1}{t + 1}\right\} \mathrm{dt} ,$

$= \frac{3}{2} \left[\ln | t | - \ln | t + 1 |\right\} ,$

$= \frac{3}{2} \ln | \frac{t}{t + 1} | , \mathmr{and} , \because , t = {x}^{2} ,$

$I = \frac{3}{2} \ln \left\{{x}^{2} / \left({x}^{2} + 1\right)\right\} + C , \mathmr{and} ,$

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

Enjoy Maths.!