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

May 9, 2018

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

#### Explanation:

We seek:

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

We may be tempted to factorise the given factors in the denominator into complex factors. Although this would work it is inadvisable as its cumbersome and possibly beyond the ability of students studying partial fractions.

Therefore we leave the factors and the partial fraction decomposition of the integrand would be of the form:

$\frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)} \equiv \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 2}$

Where $A , B , C , D$ are constant coefficients to be determined. Which leads to:

${x}^{3} + {x}^{2} + 2 x + 1 \equiv \left(A x + B\right) \left({x}^{2} + 2\right) + \left(C x + D\right) \left({x}^{2} + 1\right)$

And by comparing coefficients, we get:

$C o e f f \left({x}^{3}\right) : A + B = 1$
$C o e f f \left({x}^{2}\right) : B + D = 1$
$C o e f f \left({x}^{1}\right) : 2 A + C = 2$
$C o e f f \left({x}^{0}\right) : 2 B + D = 1$

And solving these simultaneously we obtain:

$A = 1 , B = 0 , C = 0 , D = 1$

Allowing us to write:

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

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

Consider the first integral:

${I}_{1} = \int \setminus \frac{x}{{x}^{2} + 1} \setminus \mathrm{dx}$

We can perform a substitution:

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

so that:

${I}_{1} = \int \setminus \frac{\frac{1}{2}}{u} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{1}{2} \ln | u | + C$

And restoration of the substitution gives:

${I}_{1} = \frac{1}{2} \ln | {x}^{2} + 1 | + C$
$\setminus \setminus \setminus = \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$

Now, Consider the second integral:

${I}_{2} = \int \setminus \frac{1}{{x}^{2} + 2} \setminus \mathrm{dx}$

We can perform a substitution:

$\sqrt{2} u = x \implies \sqrt{2} \frac{\mathrm{du}}{\mathrm{dx}} = 1$

so that:

${I}_{2} = \int \setminus \frac{1}{2 {u}^{2} + 2} \setminus \sqrt{2} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{\sqrt{2}}{2} \setminus \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{\sqrt{2}}{2} \arctan \left(u\right) + C '$

And restoration of the substitution gives:

${I}_{2} = \frac{\sqrt{2}}{2} \arctan \left(\frac{x}{\sqrt{2}}\right) + C '$

Combining these results, we get:

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

May 9, 2018

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

#### Explanation:

$\text{The answer given by"color(blue)" Steve M.}$ is best and applied to this type of all question.But in INDIA we give some tips to students about time saving methods. Students can easily obtain partial fractions.

We have,

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

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

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

=int(x(x^2+2))/((x^2+1)(x^2+2))dx+int(x^2+1)/((x^2+1) (x^2+2))dx

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

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

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

$= \frac{1}{2} \ln \left({x}^{2} + 1\right) + \frac{\sqrt{2}}{2} {\tan}^{-} 1 \left(\frac{x}{\sqrt{2}}\right) + c$
Note:

This method can use for a question of the form:

$I = \int \frac{{x}^{3} + {x}^{2} + \textcolor{red}{m} x + \textcolor{red}{n}}{\left({x}^{2} + \textcolor{red}{m}\right) \left({x}^{2} + \textcolor{red}{n}\right)} \mathrm{dx} , w h e r e , \textcolor{red}{m} , \textcolor{red}{n} \in \mathbb{Z}$

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

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

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

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

$= \frac{1}{2} \ln \left({x}^{2} + n\right) + \frac{1}{\sqrt{m}} {\tan}^{-} 1 \left(\frac{x}{\sqrt{m}}\right) + c$

Teachers can use this form to give several questions for practice by choosing different values of m and n.