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Evaluate using integration by parts: integral {5 x^3 + 8 x^2 - 10 x - 39} / {(x - 1)^2 (x + 2)} dx.

Question:

Evaluate using integration by parts:

{eq}\displaystyle \int \dfrac {5 x^3 + 8 x^2 - 10 x - 39} {(x - 1)^2 (x + 2)}\ dx {/eq}.

Integration by Partial Fraction:

The different types of partial fractions are listed below:

Linear Partial Fraction: {eq}\displaystyle nu+m, {/eq} its partial fractions are given by {eq}\displaystyle \frac{N}{nu+m}. {/eq}

Repeated Linear Factor: {eq}\displaystyle (nu+m)^2, {/eq} its partial fractions are given by {eq}\displaystyle \frac{N}{nu+m}+\frac{M}{(nu+m)^2}. {/eq}

A quadratic factor:{eq}\displaystyle nu^2+mu+o, {/eq} its partial fractions are given by {eq}\displaystyle \frac{Nu+M}{nu^2+mu+o}. {/eq}

Historically important notations we used to evaluate the integration:

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

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

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

5. The exponent property: {eq}\displaystyle \frac{1}{u^2}=u^{-2}. {/eq}

Answer and Explanation:

We have to solve the integration of $$\displaystyle I = \int \dfrac {5 x^3 + 8 x^2 - 10 x - 39} {(x - 1)^2 (x + 2)} dx $$

Expand:

$$\displaystyle = \int \frac{5x^3+8x^2-10x-39}{x^3-3x+2} dx $$

Apply the long division.

Step 1. Divide the leading coefficients of the numerator {eq}5x^3+8x^2-10x-39 {/eq} and the divisor {eq}x^3-3x+2. {/eq}

Then we get Quotient {eq}=5 {/eq} and Remainder {eq}=8x^2+5x-49. {/eq}

Therefore, {eq}5+\frac{8x^2+5x-49}{x^3-3x+2}. {/eq}

So, we get integration in the form as

$$\begin{align*} \displaystyle &=\int 5+\frac{8x^2+5x-49}{x^3-3x+2} dx\\ \displaystyle &= \int 5+\frac{8x^2+5x-49}{\left(x-1\right)^2\left(x+2\right)} dx \end{align*} $$

Now, we are going to find its partial fractions.

{eq}\displaystyle \frac{8x^2+5x-49}{\left(x-1\right)^2\left(x+2\right)}=\frac{a_0}{x-1}+\frac{a_1}{\left(x-1\right)^2}+\frac{a_2}{x+2} {/eq}

Simplify:

{eq}\displaystyle 8x^2+5x-49=a_0\left(x-1\right)\left(x+2\right)+a_1\left(x+2\right)+a_2\left(x-1\right)^2 {/eq}

By using its root 1, we get the value of {eq}a_1=-12 {/eq} and another root -2, we get the value of {eq}a_2=-3. {/eq}

Put in the value of {eq}a_1 {/eq} and {eq}a_2. {/eq}

{eq}\displaystyle 8x^2+5x-49=a_0\left(x-1\right)\left(x+2\right)+\left(-12\right)\left(x+2\right)+\left(-3\right)\left(x-1\right)^2 {/eq}

Expand:

{eq}\displaystyle 8x^2+5x-49=a_0x^2-3x^2+a_0x-6x-2a_0-27 {/eq}

Group the elements accourding to powers of x.

{eq}\displaystyle 8x^2+5x-49=x^2\left(a_0-3\right)+x\left(a_0-6\right)+\left(-2a_0-27\right) {/eq}

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

{eq}\displaystyle -2a_0-27=-49. {/eq}

Now, solving above equations we get the value of {eq}a_0=11. {/eq}

So, partial fractions are given by

$$\displaystyle = \int 5-\frac{3}{x+2}+\frac{11}{x-1}-\frac{12}{\left(x-1\right)^2}dx $$

Apply the sum rule.

$$\displaystyle = \int \:5dx-\int \frac{3}{x+2}dx+\int \frac{11}{x-1}dx-\int \frac{12}{\left(x-1\right)^2}dx $$

Eject the constant out.

$$\displaystyle = \int 5dx-3 \cdot \int \frac{1}{x+2}dx11 \cdot +\int \frac{1}{x-1}dx-12 \cdot \int \frac{1}{\left(x-1\right)^2}dx $$

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

$$\displaystyle = \int 5dx-3 \cdot \int \frac{1}{x+2}dx11 \cdot +\int \frac{1}{x-1}dx-12\cdot \int \frac{1}{u^2}du $$

Use the exponent property.

$$\displaystyle = \int 5dx-3 \cdot \int \frac{1}{x+2}dx11 \cdot +\int \frac{1}{x-1}dx-12\cdot \int u^{-2}du $$

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

$$\displaystyle = 5x-3\ln \left|x+2\right|+11\ln \left|x-1\right|-12\cdot \frac{u^{-2+1}}{-2+1}+C $$

Put in the value of u.

$$\displaystyle = 5x-3\ln \left|x+2\right|+11\ln \left|x-1\right|-12\cdot \frac{\left(x-1\right)^{-2+1}}{-2+1}+C $$

Simplify:

$$\displaystyle = 5x-3\ln \left|x+2\right|+11\ln \left|x-1\right|+\frac{12}{x-1}+C $$

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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