Evaluate the indefinite integral. \int \frac{x^4+5x^2-7x+9}{x^3+x^2+3x-5} dx

Question:

Evaluate the indefinite integral.

{eq}\int \frac{x^4+5x^2-7x+9}{x^3+x^2+3x-5} dx {/eq}

Integration by Partial Fraction:

The conversion process of radical fraction to a linear faction is called the partial fraction method. For example, we have the integration

{eq}\displaystyle \int \frac{z(t)}{(t+a)(t+b)} dt = \int \frac{A}{t+a}+\frac{B}{t+b}. {/eq}

Some important mathematical formulas:

1. The sum rule: {eq}\displaystyle \int f\left(u\right)\pm g\left(u\right)du=\int f\left(u\right)du\pm \int g\left(u\right)du. {/eq}

2. The power rule: {eq}\displaystyle \int u^adu=\frac{u^{a+1}}{a+1}, \quad a\ne -1. {/eq}

3. Integration of a constant: {eq}\displaystyle \int adu=au. {/eq}

4. Common integration: {eq}\displaystyle \int \frac{1}{u}du=\ln \left(\left|u\right|\right). {/eq}

5. Common integration: {eq}\displaystyle \int \frac{1}{u^2+1}du=\arctan \left(u\right). {/eq}

6. Eject the constant out: {eq}\displaystyle \int a\cdot f\left(u\right)du=a\cdot \int f\left(u\right)du. {/eq}

Answer and Explanation:

The task is to solve the integration of $$\displaystyle I = \int \frac{x^4+5x^2-7x+9}{x^3+x^2+3x-5} dx $$

Apply the long division.

Step 1. Divide the leading coefficients of the numerator {eq}x^4+5x^2-7x+9 {/eq} and the divisor {eq}x^3+x^2+3x-5. {/eq}

Then we get Quotient {eq}=x {/eq} and Remainder {eq}=-x^3+2x^2-2x+9. {/eq}

Therefore, {eq}x+\frac{-x^3+2x^2-2x+9}{x^3+x^2+3x-5}. {/eq}

Step 2. Divide the leading coefficients of the numerator {eq}-x^3+2x^2-2x+9 {/eq} and the divisor {eq}x^3+x^2+3x-5. {/eq}

Then we get Quotient {eq}=-1 {/eq} and Remainder {eq}=3x^2+x+4. {/eq}

Therefore, {eq}-1+\frac{3x^2+x+4}{x^3+x^2+3x-5}. {/eq}

So, integral can be written as:

$$\displaystyle = \int x-1+\frac{3x^2+x+4}{x^3+x^2+3x-5} dx $$

Factor:

$$\displaystyle = \int x-1+\frac{3x^2+x+4}{\left(x-1\right)\left(x^2+2x+5\right)} dx $$

Now, we are going to find its partial fractions.

{eq}\displaystyle \frac{3x^2+x+4}{\left(x-1\right)\left(x^2+2x+5\right)}=\frac{a_0}{x-1}+\frac{a_2x+a_1}{x^2+2x+5} {/eq}

Simplify:

{eq}\displaystyle 3x^2+x+4=a_0\left(x^2+2x+5\right)+\left(a_2x+a_1\right)\left(x-1\right) {/eq}

By using its root 1, we get the value of {eq}a_0 = 1. {/eq}

Put in the value of {eq}a_0. {/eq}

{eq}\displaystyle 3x^2+x+4=1\cdot \left(x^2+2x+5\right)+\left(a_2x+a_1\right)\left(x-1\right) {/eq}

Expand:

{eq}\displaystyle 3x^2+x+4=x^2+2x+5+a_2x^2-a_2x+a_1x-a_1 {/eq}

Group the elements accourding to powers of x.

{eq}\displaystyle 3x^2+1\cdot \:x+4=x^2\left(a_2+1\right)+x\left(a_1+2-a_2\right)+\left(-a_1+5\right) {/eq}

Equate the coefficients of similar terms on both sides to create a list of equations:

{eq}\displaystyle 5-a_1=4\\ \displaystyle 2-a_2+a_1=1\\ \displaystyle 1+a_2=3. {/eq}

Now, solving above equations we get the value of {eq}a_2=2 {/eq} and {eq}a_1=1. {/eq}

So, partial fractions are given by

$$\displaystyle = \int x-1+\frac{1}{x-1}+\frac{2x+1}{x^2+2x+5}dx $$

Apply the sum rule.

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2x+1}{x^2+2x+5}dx $$

Complete the square.

$$\begin{align*} \displaystyle &= \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2x+1}{x^2+2x+5+1^2-1^2}dx\\ \displaystyle &= \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2x+1}{\left(x+1\right)^2+5-1^2}dx\\ \displaystyle &= \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2x+1}{\left(x+1\right)^2+4}dx \end{align*} $$

Take {eq}u=x+1 \Rightarrow du = dx. {/eq}

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2u-1}{u^2+4}du $$

Expand:

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2u}{u^2+4}-\frac{1}{u^2+4}du $$

Apply the sum rule.

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+\int \frac{2u}{u^2+4}du-\int \frac{1}{u^2+4}du $$

Eject the constant out.

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+2 \cdot \int \frac{u}{u^2+4}du-\int \frac{1}{u^2+4}du $$

Take {eq}v=u^2+4 \Rightarrow dv=2u du\\ u=2w \Rightarrow du = 2 dw. {/eq}

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+2\cdot \int \frac{1}{2v}dv-\int \frac{1}{2\left(v^2+1\right)}dv $$

Eject the constant out.

$$\displaystyle = \int xdx-\int 1dx+\int \frac{1}{x-1}dx+2\cdot \frac{1}{2}\cdot \int \frac{1}{v}dv-\frac{1}{2}\cdot \int \frac{1}{w^2+1}dw $$

Use the power rule, integration of a constant and common integration.

$$\displaystyle = \frac{x^2}{2}-x+\ln \left|x-1\right|+2\cdot \frac{1}{2}\ln \left|v\right|-\frac{1}{2}\arctan \left(w\right)+C $$

Put in the value of v and w.

$$\displaystyle = \frac{x^2}{2}-x+\ln \left|x-1\right|+2\cdot \frac{1}{2}\ln \left|u^2+4\right|-\frac{1}{2}\arctan \left(\frac{u}{2}\right)+C $$

Put in the value of u.

$$\displaystyle = \frac{x^2}{2}-x+\ln \left|x-1\right|+\ln \left|\left(x+1\right)^2+4\right|-\frac{1}{2}\arctan \left(\frac{x+1}{2}\right)+C $$

Simplify:

$$\displaystyle = \frac{x^2}{2}-x+\ln \left|x-1\right|+\ln \left|x^2+2x+5\right|-\frac{1}{2}\arctan \left(\frac{1}{2}\left(x+1\right)\right)+C $$

Where C is constant of the integration.


Learn more about this topic:

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How to Integrate Functions With Partial Fractions

from Math 104: Calculus

Chapter 13 / Lesson 9
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