Solve the integral \int \frac{x+3}{(x-1)^3} \mathrm{d}x .

Question:

Solve the integral {eq}\int \frac{x+3}{(x-1)^3} \mathrm{d}x {/eq}.

Solve Integrals Using Substitution:

We have to solve the given integration. We will observe that it is an example of integration substitution method. Use the integration substitution method and get the desired result.

Answer and Explanation:

We assume that the given integration will be {eq}I = \int {\frac{{\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^3}}}dx} .....\left( 1 \right). {/eq}

We will solve this integration with the help of integration substitution method and we have $$\begin{align*} u &= x - 1\\ \frac{{du}}{{dx}} &= \frac{d}{{dx}}\left( {x - 1} \right)\\ \frac{{du}}{{dx}} &= 1\\ du &= dx. \end{align*} $$

Now, the equation (1) will be $$\begin{align*} \int {\frac{{\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^3}}}dx} &= \int {\frac{{u + 4}}{{{u^3}}}dx} \\ &= \int {\left( {\frac{1}{{{u^2}}} + \frac{4}{{{u^3}}}} \right)du} \\ &= \int {\left( {\frac{1}{{{u^2}}}} \right)du} + 4 \cdot \int {\left( {\frac{1}{{{u^3}}}} \right)du} \\ &= \int {{u^{ - 2}}du} + 4 \cdot \int {{u^{ - 3}}du} &\text{use power rule}\\ &= \frac{{{u^{ - 2 + 1}}}}{{\left( { - 2 + 1} \right)}} + 4 \cdot \frac{{{u^{ - 3 + 1}}}}{{\left( { - 3 + 1} \right)}} + C &\text{C is an integral constant}\\ &= - \frac{1}{u} - \frac{2}{{{u^2}}} + C &\text{undo the substitution}\\ \boxed {\int {\frac{{\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^3}}}dx} = - \frac{1}{{\left( {x - 1} \right)}} - \frac{2}{{{{\left( {x - 1} \right)}^2}}} + C} \end{align*} $$


Learn more about this topic:

Loading...
How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
7K

Related to this Question

Explore our homework questions and answers library