# Evaluate the integrals int_C (2z^6 + 1) dz/z(z+1)^2(2z-5)^4.

## Question:

Evaluate the integrals

{eq}\int_C \frac{(2z^6 + 1) dz}{z(z+1)^2(2z-5)^4} {/eq}.

## Integration:

A mathematical quantity that represents the merge of the derivative of data or variable function is known as integration. It helps in the engineering discipline to analyze the system, and it makes it easy to calculate.

Given Data:

• The integral is: {eq}{I_o} = \int_C {\dfrac{{\left( {2{z^6} + 1} \right)dz}}{{z{{\left( {z + 1} \right)}^2}{{\left( {2z - 5} \right)}^4}}}} {/eq}

The function of the integral is

{eq}f\left( z \right) = \dfrac{{\left( {2{z^6} + 1} \right)}}{{z{{\left( {z + 1} \right)}^2}{{\left( {2z - 5} \right)}^4}}} {/eq}

The poles of the function are root of the equation

Equate the denominator of integral equal to zero to get the root of the function

{eq}\begin{align*} z{\left( {z + 1} \right)^2}{\left( {2z - 5} \right)^4} &= 0\\ z &= 0\\ z &= - 1\\ z &= \dfrac{5}{2} \end{align*} {/eq}

The residue at {eq}z = 0 {/eq} of function is

{eq}\begin{align*} {R_1} &= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{z^6} + 1} \right)}}{{{{\left( {z + 1} \right)}^2}{{\left( {2z - 5} \right)}^4}}}\\ &= \dfrac{{\left( {2{{\left( 0 \right)}^6} + 1} \right)}}{{{{\left( {0 + 1} \right)}^2}{{\left( {2\left( 0 \right) - 5} \right)}^4}}}\\ &= \dfrac{1}{{625}} \end{align*} {/eq}

The residue at {eq}z = - 1 {/eq} of function is

{eq}\begin{align*} {R_2} &= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{z^6} + 1} \right)}}{{z{{\left( {2z - 5} \right)}^4}}}\\ &= \dfrac{{\left( {2{{\left( { - 1} \right)}^6} + 1} \right)}}{{\left( { - 1} \right){{\left( {2\left( { - 1} \right) - 5} \right)}^4}}}\\ &= \dfrac{1}{{2401}} \end{align*} {/eq}

The residue at {eq}z = \dfrac{5}{2} {/eq} of function is

{eq}\begin{align*} {R_3} &= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{z^6} + 1} \right)}}{{z{{\left( {z + 1} \right)}^2}}}\\ &= \dfrac{{\left( {2{{\left( {\dfrac{5}{2}} \right)}^6} + 1} \right)}}{{\left( {\dfrac{5}{2}} \right){{\left( {\dfrac{5}{2} + 1} \right)}^2}}}\\ &= \dfrac{{\left( {\dfrac{{31314}}{{64}}} \right)}}{{\left( {\dfrac{{245}}{8}} \right)}}\\ &= 15.97 \end{align*} {/eq}

The expression for integration by residual theorem is

{eq}{I_o} = 2\pi i\left( {{R_1} + {R_2} + {R_3}} \right) {/eq}

Substitute and solve the above expression

{eq}\begin{align*} {I_o} &= 2\pi i\left( {\dfrac{1}{{625}} + \dfrac{1}{{2401}} + 15.97} \right)\\ &= 31.97\pi i \end{align*} {/eq}

Thus integral is {eq}31.97\pi i {/eq}