Blog: `int (x^3 + x^2 + 2x + 1)/((x^2 + 1)(x^2 + 2)) dx` Evaluate the integral

$\displaystyle\int\frac{{{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}}}{{{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}}}{\left.{d}{x}\right.}$

To solve, apply partial fractions decomposition.

To express the integrand as sum of proper rational expressions, set the equation as follows:

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

Multiply both sides by the LCD.

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

$\displaystyle{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}={A}{{x}}^{{3}}+{2}{A}{x}+{B}{{x}}^{{2}}+{2}{B}+{C}{{x}}^{{3}}+{C}{x}+{D}{{x}}^{{2}}+{D}$

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

For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.

x^3:

$\displaystyle{1}={A}+{C}$ (Let this be EQ1.)

x^2:

$\displaystyle{1}={B}+{D}$ (Let...

$\displaystyle\int\frac{{{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}}}{{{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}}}{\left.{d}{x}\right.}$

To solve, apply partial fractions decomposition.

To express the integrand as sum of proper rational expressions, set the equation as follows:

Multiply both sides by the LCD.

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

$\displaystyle{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}={A}{{x}}^{{3}}+{2}{A}{x}+{B}{{x}}^{{2}}+{2}{B}+{C}{{x}}^{{3}}+{C}{x}+{D}{{x}}^{{2}}+{D}$

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

For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.

x^3:

$\displaystyle{1}={A}+{C}$ (Let this be EQ1.)

x^2:

$\displaystyle{1}={B}+{D}$ (Let this be EQ2.)

$\displaystyle{2}={2}{A}+{C}$ (Let this be EQ3.)

Constant:

$\displaystyle{1}={2}{B}+{D}$ (Let this be EQ4.)

To solve for the values of A, B, C and D, isolate the C in EQ1.

$\displaystyle{1}={A}+{C}$

$\displaystyle{1}-{A}={C}$

Plug-in this to EQ3.

$\displaystyle{2}={2}{A}+{C}$

$\displaystyle{2}={2}{A}+{1}-{A}$

$\displaystyle{2}={A}+{1}$

$\displaystyle{1}={A}$

Plug-in the value of A to EQ1.

$\displaystyle{1}={A}+{C}$

$\displaystyle{1}={1}+{C}$

$\displaystyle{0}={C}$

Also, isolate the D in EQ2.

$\displaystyle{1}={B}+{D}$

$\displaystyle{1}-{B}={D}$

Plug-in this to EQ4.

$\displaystyle{1}={2}{B}+{D}$

$\displaystyle{1}={2}{B}+{1}-{B}$

$\displaystyle{1}={B}+{1}$

$\displaystyle{0}={B}$

And plug-in the value of B to EQ2.

$\displaystyle{1}={B}+{D}$

$\displaystyle{1}={0}+{D}$

$\displaystyle{1}={D}$

So the partial fraction decomposition of the integrand is:

$\displaystyle\frac{{{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}}}{{{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}}}=\frac{{{1}{x}+{0}}}{{{{x}}^{{2}}+{1}}}+\frac{{{0}{x}+{1}}}{{{{x}}^{{2}}+{2}}}=\frac{{x}}{{{{x}}^{{2}}+{1}}}+\frac{{1}}{{{{x}}^{{2}}+{2}}}$

Taking the integral of this result to:

$\displaystyle\int\frac{{{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}}}{{{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left(\frac{{x}}{{{{x}}^{{2}}+{1}}}+\frac{{1}}{{{{x}}^{{2}}+{2}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}+\int\frac{{1}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$

For the first integral, apply u-substitution method.

$\displaystyle{u}={{x}}^{{2}}+{1}$

$\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$

$\displaystyle\frac{{{d}{u}}}{{2}}={x}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{u}}\cdot{\left({d}\frac{{u}}{{2}}\right)}+\int\frac{{1}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{1}}{{u}}{d}{u}+\int\frac{{1}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{u}\right|}+\frac{{1}}{\sqrt{{2}}}{{\tan}}^{{-{1}}}{\left(\frac{{x}}{\sqrt{{2}}}\right)}+{C}$

And, substitute back $\displaystyle{u}={{x}}^{{2}}+{1}$ .

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}+\frac{{1}}{\sqrt{{2}}}{{\tan}}^{{-{1}}}{\left(\frac{{x}}{\sqrt{{2}}}\right)}+{C}$

Therefore, $\displaystyle\int\frac{{{{x}}^{{3}}+{{x}}^{{2}}+{2}{x}+{1}}}{{{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}+\frac{{1}}{\sqrt{{2}}}{{\tan}}^{{-{1}}}{\left(\frac{{x}}{\sqrt{{2}}}\right)}+{C}$ .