# Part 1 Write the form of the partial fraction decomposition of the function . Do not determine...

## Question:

Part 1 Write the form of the partial fraction decomposition of the function . Do not determine the numerical values of the coefficients.

{eq}(a) \frac{2x}{ (x + 6)(3x + 1)}\\ (b) \frac{5}{ x^3 + 10x^2 + 25x} {/eq}

Part 2 Find the area of the region under the given curve from 1 to 4.

{eq}y = \frac{x^2+2}{5x - x^2} {/eq}

Part 3 Evaluate the integral. (Remember to use absolute values where appropriate.)

{eq}\int 17x^3 - 125 dx {/eq}

## Integration.

Integration can be solved by using many methods; substitution, trigonometric substitution, integration by parts, etc.

One of the methods of solving integration is a partial method in which

the proper rational fraction of the terms of the denominator should be reduced.

The formula is:

{eq}\displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ \displaystyle\int \dfrac{1}{x}dx=ln\left | x \right |+c\\\\ {/eq}

We have to write the form of partial fraction of the given function:

Part 1a.)

{eq}\dfrac{2x}{(x+6)(3x+1)}\\\\ {/eq}

The general form of partial fraction is:

{eq}\dfrac{2x}{(x+6)(3x+1)}=\dfrac{A}{x+6}+\dfrac{B}{3x+1}\\\\ {/eq}

Part 1b.)

{eq}\dfrac{5}{x^3+10x^2+25x}\\\\ {/eq}

We may write the given integration as:

{eq}\dfrac{5}{x^3+10x^2+25x}=\dfrac{5}{x(x^2+10x+25)}\\\\ \dfrac{5}{x(x^2+10x+25)}=\dfrac{5}{x(x+5)^2}\\\\ {/eq}

Hence the general form of partial fraction is:

{eq}\dfrac{5}{x(x+5)^2}=\dfrac{A}{x}+\dfrac{B}{x+5}+\dfrac{C}{(x+5)^2}\\\\ {/eq}

Part 2.)

Given function is:

{eq}y=\dfrac{x^2+2}{5x-x^2}\\\\ {/eq}

Hence the area of the region bounded by the curve from 1 to 4 in the xy-plane is:

{eq}\displaystyle\int _1^4y\ dx\\\\ {/eq}

Hence it can be written as:

{eq}\displaystyle\int _1^4\dfrac{x^2+2}{5x-x^2}dx\\\\ {/eq}

Since here the degree of numerator is greater than denominator hence we are applying here long division rule:

{eq}\dfrac{x^2+2}{5x-x^2}=1+\dfrac{5x+2}{5x-x^2}\\\\ \dfrac{5x+2}{5x-x^2}=\dfrac{5x+2}{x(5-x)}\\\\ {/eq}

Now using the formula of partial fraction:

{eq}\dfrac{5x+2}{x(5-x)}=\dfrac{A}{x}+\dfrac{B}{5-x}\\\\ \dfrac{5x+2}{x(5-x)}=\dfrac{A(5-x)+Bx}{x(5-x)}\\\\ 5x+2=5A-Ax+Bx\\\\ {/eq}

Comparing the coefficients of x we get:

{eq}-A+B=5\,\,\,\,\,eqn(1)\\\\ {/eq}

And by the constant terms we get:

{eq}5A=2\\\\ A=\dfrac{2}{5}\\\\ {/eq}

Hence form eqn(1) we get:

{eq}B=\dfrac{27}{5}\\\\ {/eq}

Therefore integrating:

{eq}\displaystyle\int_1^4 \dfrac{x^2+2}{5x-x^2}dx=\displaystyle\int _1^41\ dx+\displaystyle\int _1^4\dfrac{\frac{2}{5}}{x}dx+\displaystyle\int _1^4\dfrac{\frac{27}{5}}{5-x}dx\\\\ {/eq}

Let:

{eq}5-x=z\\\\ -dx=dz\\\\ dx=-dz\\\\ {/eq}

Substituting for the values of limits:

Lower limits:

When {eq}x=1\\\\ {/eq}

{eq}z=4\\\\ {/eq}

Upper limit:

When {eq}x=4\\\\ {/eq}

{eq}z=1\\\\ {/eq}

Hence:

{eq}=\left [ x \right ]_1^4+\dfrac{2}{5}\left [ ln\left | x \right | \right ]_1^4+\dfrac{27}{5}\displaystyle\int _4^1\dfrac{1}{z}(-dz)\\\\ =\left [ 4-1 \right ]+\dfrac{2}{5}\left [ ln(4)-ln(1) \right ]-\dfrac{27}{5}\left [ ln\left | z \right | \right ]_4^1\\\\ =3+\dfrac{2}{5}ln(4)-\dfrac{27}{5}\left [ ln(1)-ln(4) \right ]\\\\ 3+\dfrac{2}{5}ln(4)+\dfrac{27}{5}ln(4)\\\\ {/eq}

Part 3.)

{eq}\displaystyle\int (17x^3-125)dx\\\\ =\displaystyle\int 17x^3\ dx-\displaystyle\int 125\ dx\\\\ =17\cdot \dfrac{x^4}{4}-125x+c\\\\ {/eq}

Here c denotes integration constant.