Integration Techniques, multiple pages

In this chapter, we are going to take off the gloves and work harder on the techniques used to solve integral problems.

We will still remain in the area of the indefinite integrals until we complete covering topics like substitution, trigonometrics and hyperbolic functions without omtitting our favorite integration by parts.

This chapter will involve more practice, including videos of some of the solutions.

Tha ability to solve integral equation depends very much on vision, planning and understanding of the concepts.

1.The method of substitution

We start with the method of substitution that we’ll use most of the time.

When we have $\int f(x) \, dx$ , we often replace the variable $x$ with a new variable $t$ or $u$ by means of $substitution$ .

$x=g(u)$ yielding $dx=g'(u)du$ . This technique helps us get the basic forms discussed in the previous chapter.

Let’s investigate some of the forms.

Example:

Problem 3:

Evaluate $\int (2x+5)^{4} \, dx$

Let $2x+5=u \Rightarrow 2dx=du$

We get the following form:

(1) $\begin{equation*} \begin{split} \displaystyle \int \frac{1}{2}u^{4} \, du}&={\displaystyle \frac{1}{2}\int u^{4} \, du}\\ &=\frac{1}{2}\cdot \frac{1}{4+1}u^{4+1}\\&=\frac{1}{2}\cdot \frac{1}{4+1}u^{4+1}+C\\ &=\frac{1}{2}\cdot \frac{1}{5}u^{5}\\&=\frac{1}{10}u^{5}+C \end{split} \end{equation*}$

Getting back to the initial variable:

${\displaystyle \int (2x+5)^{4} \, dx}=\frac{1}{10} (2x+5)^{5}+C$

Problem 4:

Evaluate $\int e^{-x} \, dx$

Let $-x=u \Rightarrow dx=-du$

We get the following form:

${\displaystyle -\int e^{u} \, du}=-e^{u}+C$

Getting back to the initial variable:

${\displaystyle \int (e)^{-x} \, dx}=-e^{-x}+C$

Problem 5:

Evaluate $\int \frac{x+2}{x+1} \, dx$

Let $x+2=x+1+1$

We get the following form:

(2) $\begin{equation*} \begin{split} \displaystyle \int \frac{x+2}{x+1}\, dx}&={\displaystyle \int \frac{x+1+1}{x+1}\, dx}\\ &={\displaystyle \int dx+ \int \frac{dx}{x+1}}\\ &=x+ \ln \left | x+1 \right |+C \end{split} \end{equation*}$

Finally:

${\displaystyle \int \frac{x+2}{x+1}\, dx}=x+ \ln \left | x+1 \right |+C$

Problem 6:

Evaluate $\int \frac{x^{2}-6\sqrt[3]{x}}{x} \, dx$

No substitution necessary here.

(3) $\begin{equation*} \begin{split} \displaystyle \int \frac{x^{2}-6\sqrt[3]{x}}{x} \, dx}&={\displaystyle \int \frac{x^{2}}{x}\, dx}-{\displaystyle 6\int \frac {\sqrt[3]{x}}{x}\, dx}\\ &= {\displaystyle \int x\, dx}-{\displaystyle 6\int \frac {x^{\frac{1}{3}}}{x^{\frac{3}{3}}}\, dx}\\ &={\displaystyle \int x\, dx}-{\displaystyle 6\int x^{\frac{1}{3}} \cdot x^{\frac{-3}{3}}\, dx}\\ &={\displaystyle \int x\, dx}-{\displaystyle 6\int x^{\frac{-2}{3}}\, dx}\\ &={\displaystyle \frac{x^{2}}{2}-6\frac{x^{\frac{1}{3}}}{\frac{1}{3}}}+C \\ &={\displaystyle \frac{x^{2}}{2}-6\frac{x^{\frac{1}{3}}}{\frac{1}{3}} }+C\\ &={\displaystyle \frac{x^{2}}{2}-18 x^{\frac{1}{3}}}+C \\ &={\displaystyle \frac{x^{2}}{2}-18 \sqrt[3]{x} }+C \end{split} \end{equation*}$

Finally:

${\displaystyle \int \frac{x^{2}-6\sqrt[3]{x}}{x} \, dx}={\displaystyle \frac{x^{2}}{2}-18 \sqrt[3]{x} }+C$

Problem 7

Evaluate: $\int \frac{2x-3}{4x^{2}-11}\, dx$

We know that $\frac{d}{dx}(4x^{2}-11)=8x$
From $(2x-3)$
$2x-3=\frac{1}{4}(8x-12)$

We can write

(4) $\begin{equation*} \begin{split} \displaystyle \int \frac{2x-3}{4x^{2}-11}\, dx}&={\displaystyle \frac{1}{4} \int \frac{8x-12}{4x^{2}-11}\, dx}\\ &={\displaystyle \frac{1}{4} \int\frac{8x}{4x^{2}-11}\, dx -\frac{12}{4} \int\frac{dx}{4x^{2}-11}}\\ &={\displaystyle \frac{1}{4} \int\frac{8x}{4x^{2}-11}\, dx -\frac{3}{2} \int\frac{2dx}{(2x)^{2}-(\sqrt{11})^{2}}}\\ &={\displaystyle \frac{1}{4} \ln \left |4x^{2}-11 \right | -\frac{3}{2}\cdot \frac{1}{2 \sqrt{11}} \ln \left|\frac{2x-\sqrt{11}}{2x+\sqrt{11}}\right|+C}\\ &={\displaystyle \frac{1}{4} \ln \left |4x^{2}-11 \right | -\frac{3\sqrt{11}}{44} \ln \left|\frac{2x-\sqrt{11}}{2x+\sqrt{11}}\right|+C} \end{split} \end{equation*}$

Finally we have:

$\int \frac{2x-3}{4x^{2}-11}\, dx}= {\displaystyle \frac{1}{4} \ln \left |4x^{2}-11 \right | -\frac{3\sqrt{11}}{44} \ln \left|\frac{2x-\sqrt{11}}{2x+\sqrt{11}}\right|+C$

Problem 8
Evaluate: $\int \frac{x+3}{4x^{2}+12x}\, dx$

$\int \frac{x+3}{4x^{2}+24x}\, dx={\displaystyle \frac{1}{4}\int \frac{x+3}{x^{2}+6x}\, dx}$

Let $u=x^{2}+6x$

We get:

$du=(2x+6)dx \Rightarrow dx=\frac{du}{2(x+3)}$

(5) $\begin{equation*} \begin{split} \displaystyle \int \frac{x+3}{4x^{2}+24x}\, dx}&={\displaystyle \frac{1}{4} \int \frac{(x+3)du}{2(x+3)u}}\\ &={\displaystyle \frac{1}{8} \int \frac{du}{u}}\\ &= {\displaystyle \frac{1}{8} \ln | u |}+C \end{split} \end{equation*}$

Now back to $x$ :

Finally:
${\displaystyle \int \frac{x+3}{4x^{2}+24x}\, dx}=\frac{1}{8} \ln |x^{2}+6x|+C$

Problem 9
Evaluate: $\int \frac{x^{5}-5x^{4}+3x-2}{x}\, dx$

(6) $\begin{equation*} \begin{split} \displaystyle \int \frac{x^{5}-5x^{4}+3x-2}{x}\, dx}&={\displaystyle \int \frac{x^{5}}{x}\, dx}-{\displaystyle 5 \int \frac{x^{4}}{x}\, dx}+{\displaystyle 3 \int \frac{x}{x}\,dx}-{\displaystyle 2\int \frac{dx}{x}} \\ &={\displaystyle \int x^{4}\, dx}-{\displaystyle 5 \int x^{3}\, dx}+{\displaystyle 3 \int dx}-{\displaystyle 2 \int \frac{dx}{x}} \\ &=\frac{x^{5}}{5}-5\cdot \frac{x^{4}}{4}+3x-2 \ln |x| +C \\&=\frac{1}{5}x^{5}-\frac{5}{4}x^{4}+3x-2 \ln |x| +C \end{split} \end{equation*}$

Finally:

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

Problem 10
Evaluate: $\int \frac{x^{2}}{\sqrt[4]{x^{3}+2}}\, dx$

We’ll use substitution:

Let $u=x^{3}+2 \Rightarrow du=3x^{2}dx$

(7) $\begin{equation*} \begin{split} \displaystyle \int \frac{x^{2}}{\sqrt[4]{x^{3}+2}}\, dx}&={\displaystyle \frac{1}{3} \int \frac{3x^{2}}{\sqrt[4]{x^{3}+2}}\, dx}\\ &={\displaystyle \frac{1}{3} \int \frac{du}{\sqrt[4]{u}}}\\&={\displaystyle \frac{1}{3} \int u^{-\frac{1}{4}}du}\\ &=\frac{1}{3}\cdot \frac{u^{-\frac{1}{4}+\frac{4}{4}}}{-\frac{1}{4}+\frac{4}{4}}+C\\&=\frac{4}{9}u^{\frac{3}{4}}+C \end{split} \end{equation*}$

Back to $x$

(8) $\begin{equation*} \begin{split} \displaystyle \int \frac{x^{2}}{\sqrt[4]{x^{3}+2}}\, dx} &=\frac{4}{9}(x^{3}+2)^{\frac{3}{4}}+C\\ &=\frac{4}{9}\sqrt[4]{(x^{3}+2)^{3}}+C \end{split} \end{equation*}$

Finally:

${\displaystyle \int \frac{x^{2}}{\sqrt[4]{x^{3}+2}}\, dx}=\frac{4}{9}\sqrt[4]{(x^{3}+2)^{3}}+C$

Problem 11
Evaluate: $\int 3x \sqrt{1-2x^{2}}\, dx$

Let $u=1-2x^{2} \Rightarrow du=-4xdx$

We can now write:

(9) $\begin{equation*} \begin{split} \displaystyle \int 3x \sqrt{1-2x^{2}}\, dx}&={\displaystyle -\frac{3}{4}\int (-4x)\sqrt{1-2x^{2}}\, dx}\\ &={\displaystyle -\frac{3}{4}\int \sqrt{u}\, du}\\&= {\displaystyle -\frac{3}{4}\int u^{\frac{1}{2}}\, du}\\ &=-\frac{3}{4}\cdot \frac{1}{\frac{1}{2}+\frac{2}{2}}u^{\frac{1}{2}+\frac{2}{2}}+C\\ &=-\frac{3}{4}\cdot \frac{1}{\frac{3}{2}}u^{\frac{3}{2}}+C\\&=-\frac{1}{2}u\sqrt{u}+C \end{split} \end{equation*}$

Back to $x$

${\displaystyle \int 3x \sqrt{1-2x^{2}}\, dx}=-\frac{1}{2}(1-2x^{2})\sqrt{1-2x^{2}}+C$

Finally:

$\mathbf{{\displaystyle \int 3x \sqrt{1-2x^{2}}\, dx}=-\frac{1}{2}(1-2x^{2})\sqrt{1-2x^{2}}+C}$

Problem 12
Evaluate: $\int \frac{x^{2}+8x}{(x+4)^{2}}\, dx$

(10) $\begin{equation*} \begin{split} \displaystyle \int \frac{x^{2}+8x}{(x+4)^{2}}\, dx}&={\displaystyle \int \frac{x^{2}+8x+16-16}{(x+4)^{2}}\, dx}\\ &={\displaystyle \int \frac{(x+4)^{2}-16}{(x+4)^{2}}\, dx}\\ &= {\displaystyle \int \frac{(x+4)^{2}}{(x+4)^{2}}\, dx}-{\displaystyle 16 \int \frac{dx}{(x+4)^{2}} }\\ &={\displaystyle \int dx }-{\displaystyle 16 \int (x+4)^{-2}\, dx } \end{split} \end{equation*}$

But we notice that $\frac{d}{dx}(x+4)=dx$

We get:

(11) $\begin{equation*} \begin{split} \displaystyle \int \frac{x^{2}+8x}{(x+4)^{2}}\, dx}&={\displaystyle \int dx }-{\displaystyle 16 \int (x+4)^{-2}\, dx }\\ &=x-16 \cdot \frac{1}{-2+1}(x+4)^{-2+1}+C\\&=x+16(x+4)^{-1}\\ &=x+\frac{16}{x+4}+C \end{split} \end{equation*}$

This result can be broken to show how dangerous it is to dismiss results of indefinite integrals involving constants.

(12) $\begin{equation*} \begin{split} {\displaystyle \int \frac{x^{2}+8x}{(x+4)^{2}}\, dx}&=x+\frac{16}{x+4}+C\\ &=\frac{x^{2}+4x+16}{x+4}+C\\ &=\frac{x^{2}}{x+4}+4\frac{x+4}{x+4}\\ &=\frac{x^{2}}{x+4}+4+C\\ &=\frac{x^{2}}{x+4}+C_{1} \end{split} \end{equation*}$