# Use improper fractions to integrate: \int \frac{x^3 + 10x^2 + 3x + 36}{(x + 1)(x^2 + 4)^2}

## Question:

Use improper fractions to integrate:

{eq}\int \frac{x^3 + 10x^2 + 3x + 36}{(x + 1)(x^2 + 4)^2} {/eq}

## Improper Fraction.

It is a part of the partial fraction.

This defines the function in which numerator is greater than denominator or either equal to the denominator.

Formula of integration is:

{eq}\displaystyle\int \dfrac{1}{x}dx=ln\left | x \right |+c\\\\ \displaystyle\int \dfrac{1}{x^2+a^2}dx=\dfrac{1}{a}tan^{-1}\left ( \dfrac{x}{a} \right )+c\\\\ {/eq}

We have to solve here given improper fraction:

{eq}\displaystyle \int \frac{x^3 + 10x^2 + 3x + 36}{(x + 1)(x^2 + 4)^2}dx\\\\ {/eq}

Applying partial fraction decomposition method:

{eq}\frac{x^3 + 10x^2 + 3x + 36}{(x + 1)(x^2 + 4)^2}=\dfrac{Ax+B}{x^2+4}+\dfrac{Cx+D}{(x^2+4)^2}+\dfrac{E}{x+1}\\\\ \frac{x^3 + 10x^2 + 3x + 36}{(x + 1)(x^2 + 4)^2}=\dfrac{(Ax+B)(x^2+4)(x+1)+(Cx+D)(x+1)+E(x^2+4)^2}{(x+1)(x^2+4)^2}\\\\ x^3+10x^2+3x+36=x^4(A+E)+x^3(A+B)+x^2(4A+B+C+8E)+x(4A+4B+C+D)+4B+D+16E\\\\ {/eq}

Comparing the coefficients of {eq}x^4 {/eq} from both side of the above equation we get:

{eq}A+E=0\,\,\,\,\,eqn(1)\\\\ {/eq}

Comparing the coefficients of {eq}x^3 {/eq} from both side of the above equation we get:

{eq}A+B=1\,\,\,\,\,eqn(2)\\\\ {/eq}

Comparing the coefficients of {eq}x^2 {/eq} from both side of the above equation we get:

{eq}4A+B+C+8E=10\,\,\,\,\,eqn(3)\\\\ {/eq}

Comparing the coefficients of {eq}X {/eq} from both side of the above equation we get:

{eq}4A+4B+C+D=3\,\,\,\,\,eqn(4)\\\\ {/eq}

And by comparing the constant terms we get:

{eq}4B+D+16E=36\,\,\,\,\,eqn(5)\\\\ {/eq}

Hence solving eqn 1,2,3,4 and 5 we get:

{eq}A=\dfrac{-42}{25}\\\\ B=\dfrac{67}{25}\\\\ C=\dfrac{3}{5}\\\\ D=\dfrac{-8}{5}\\\\ E=\dfrac{42}{25}\\\\ {/eq}

Substituting this value and now integrating the fraction:

{eq}\displaystyle\int \dfrac{x^3+10x^2+3x+36}{(x+1)(x^2+4)^2}dx=\displaystyle\int \dfrac{\frac{-42x}{25}+\frac{67}{25}}{x^2+4}dx+\displaystyle\int \dfrac{\frac{3x}{5}-\frac{8}{5}}{x^2+4}dx+\displaystyle\int \dfrac{\frac{42}{25}}{x+1}dx\\\\ =\displaystyle\int \dfrac{\frac{-42x}{25}}{x^2+4}dx+\dfrac{67}{25}\displaystyle\int \dfrac{dx}{x^2+(2)^2}+\dfrac{3}{5}\displaystyle\int \dfrac{x}{x^2+4}dx-\dfrac{8}{5}\displaystyle\int \dfrac{dx}{x^2+4}+\dfrac{42}{25}\displaystyle\int \dfrac{dx}{x+1}\\\\ {/eq}

Let:

{eq}x^+4=z\\\\ 2x\ dx=dz\\\\ {/eq}

Or,

{eq}x\ dx=\dfrac{dz}{2}\\\\ {/eq}

{eq}=\dfrac{-42}{25}\displaystyle\int \dfrac{1}{z}\cdot \dfrac{dz}{2}+\dfrac{67}{25}\cdot \dfrac{1}{2}tan^{-1}\left ( \dfrac{x}{2} \right )+\dfrac{3}{5}\displaystyle\int \dfrac{1}{z}\cdot \dfrac{dz}{2}-\dfrac{8}{5}\cdot \dfrac{1}{2}tan^{-1}\left ( \dfrac{x}{2} \right )+\dfrac{42}{25}ln\left | x+1 \right |+c\\\\ =\dfrac{-42}{50}ln\left | z \right |+\dfrac{67}{50}tan^{-1}\left ( \dfrac{x}{2} \right )+\dfrac{3}{10}ln\left | z \right |-\dfrac{8}{10}tan^{-1}\left ( \dfrac{x}{2} \right )+\dfrac{42}{25}ln\left | x+1 \right |+c\\\\ {/eq}

Substituting back the value of z we get:

{eq}=\dfrac{-42}{50}ln\left | x^2+4 \right |+\dfrac{67}{50}tan^{-1}\left ( \dfrac{x}{2} \right )+\dfrac{3}{10}ln\left | x^2+4 \right |-\dfrac{8}{10}tan^{-1}\left ( \dfrac{x}{2} \right )+\dfrac{42}{25}ln\left | x+1 \right |+c\\\\ {/eq}

Here c represent the integration constant.