Evaluate the integral. irreducible quadratic factor. 1. \int \frac{7x^{2} + x + 54}{x^{3} +...

Question:

Evaluate the integral. irreducible quadratic factor.

1. {eq}\int \frac{7x^{2} + x + 54}{x^{3} + 9x} {/eq}

2. {eq}\int \frac{2x^{2} + x + 3}{\left ( x^{2} + 5 \right )\left ( x - 7 \right )} {/eq}

Integration.

If the given integrands may difficult and cannot be solved using substitution, integration by parts, then we can apply the partial rule.

Here every polynomial with real variables can be decomposing into quadratic factors.

The formula 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}

Answer and Explanation:

We have to calculate the given integrals:

Part 1.)

{eq}\displaystyle\int \dfrac{7x^2+x+54}{x^3+9x}dx\\\\ \dfrac{7x^2+x+54}{x^3+9x}=\dfrac{7x^2+x+54}{x(x^2+9)}\\\\ {/eq}

Applying partial fraction method:

{eq}\dfrac{7x^2+x+54}{x(x^2+9)}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+9}\\\\ \dfrac{7x^2+x+54}{x(x^2+9)}=\dfrac{A(x^2+9)+(Bx+C)x}{x(x^2+9)}\\\\ 7x^2+x+54=x^2A+x^2B+xC+9A\\\\ {/eq}

Comparing the coefficients of {eq}x^2 {/eq} we get:

{eq}A+B=7\,\,\,\,\,eqn(1)\\\\ {/eq}

Comparing the coefficients of {eq}x {/eq} we get:

{eq}C=1\\\\ {/eq}

And by comparing the constant terms we get:

{eq}9A=54\\\\ A=6\\\\ {/eq}

Therefore form eqn(1) we get:

{eq}B=1\\\\ {/eq}

Now integrating:

{eq}\displaystyle\int \dfrac{7x^2+x+54}{x^3+9x}dx=\displaystyle\int \dfrac{1}{x}dx+\displaystyle\int \dfrac{x+1}{x^2+9}dx\\\\ =ln\left | x \right |+\displaystyle\int \dfrac{x}{x^2+9}dx+\displaystyle\int \dfrac{1}{x^2+9}dx\\\\ {/eq}

Let:

{eq}x^2+9=z\\\\ 2x\ dx=dz\\\\ x\ dx=\dfrac{dz}{2}\\\\ {/eq}

{eq}=ln\left | x \right |+\displaystyle\int \dfrac{1}{z}\cdot \dfrac{dz}{2}+\displaystyle\int \dfrac{1}{x^2+3^2}dx\\\\ =ln\left | x \right |+\dfrac{1}{2}ln\left | z \right |+\dfrac{1}{3}tan^{-1}\left ( \dfrac{x}{3} \right )+c\\\\ =ln\left | x \right |+\dfrac{1}{2}ln\left | x^2+9 \right |+\dfrac{1}{3}tan^{-1}\left ( \dfrac{x}{3} \right )+c\\\\ {/eq}

Here c is integration constant.

Part 2.)

{eq}\displaystyle\int \dfrac{2x^2+x+3}{(x^2+5)(x-7)}dx\\\\ {/eq}

Using the rule of partial fraction we get:

{eq}\dfrac{2x^2+x+3}{(x-7)(x^2+5)}dx=\dfrac{A}{x-7}+\dfrac{Bx+C}{x^2+5}\\\\ \dfrac{2x^2+x+3}{(x-7)(x^2+5)}dx=\dfrac{A(x^2+5)+(Bx+C)(x-7)}{(x-7)(x^2+5)}\\\\ 2x^2+x+3=x^2A+x^2B-7xB+xC+5A-7C\\\\ {/eq}

Comparing the coefficients of {eq}x^2 {/eq} we get:

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

Comparing the coefficients of {eq}x {/eq} we get:

{eq}-7B+C=-1\,\,\,\,\,eqn(2)\\\\ {/eq}

And by comparing the constant terms we get:

{eq}5A-7C=3\,\,\,\,\,eqn(3)\\\\ {/eq}

Solving eqn 1,2 and 3 we get:

{eq}A=2\\\\ B=0\\\\ C=1\\\\ {/eq}

Hence:

{eq}\displaystyle\int \dfrac{2x^2+x+3}{(x^2+5)(x-7)}dx=\displaystyle\int \dfrac{2}{x-7}dx+\displaystyle\int \dfrac{0+1}{x^2+5}dx\\\\ =2ln\left | x-7 \right |+\displaystyle\int \dfrac{1}{x^2+(\sqrt 5)^2}dx\\\\ =2ln\left | x-7 \right |+\dfrac{1}{\sqrt 5}tan^{-1}\left ( \dfrac{x}{\sqrt 5} \right )+c\\\\ {/eq}

Here c is integration constant.


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