In the previous section, we saw integration by method of inspection. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved example 23.2
Find the following integrals:
$(i)~\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}~~~~(ii)~\int {\left[x^{2/3} \,+\,1\right]dx}$
$(iii)~\int{\left[x^{3/2} \,+\, 2 e^x
\,-\, \frac{1}{x}\right] \,
dx}~~~~(iv)~\int{\left[\frac{4}{1+x^2}\right] \, dx}$
Solution:
Part (i):
1. We want $\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}$
• This can be written as:
$\int{\left[x~-~\frac{1}{x^2} \right] \, dx}$
⇒ $\int{\left[x~-~x^{-2} \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x \right]\,dx}~-~\int{\left[x^{-2} \right] \, dx}$
• We search for F and find that:
♦ x, is the derivative of $\frac{x^2}{2}$
♦ x−2, is the derivative of $\frac{x^{-1}}{-1}$ which is −(1/x)
2. So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}} & {~=~} &{\int{\left[x \right]\,dx}~-~\int{\left[x^{-2} \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left[\frac{x^2}{2}\,+\,C_1 \right] ~-~\left[\frac{-1}{x}\,+\,C_2 \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{x^2}{2}\,+\, \frac{1}{x} \,+\,C_1 \,-\,C_2} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{x^2}{2}\,+\, \frac{1}{x} \,+\,C} \\
\end{array}$
•
This is the general form of the required anti derivative.
◼ Remarks:
4 (magenta color): Addition/subtraction of any two real numbers C1 and C2 will give another real number which can be denoted in general as C.
• Or we can write:
$\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}~=~\frac{x^2}{2}\,+\, \frac{1}{x}$ is an anti derivative of $\frac{x^3 - 1}{x^2}$.
Part (ii):
1. We want $\int{\left[x^{2/3} \,+\,1 \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x^{2/3} \right]\,dx}~+~\int{\left[1 \right] \, dx}$
• We search for F and find that:
♦ x2/3, is the derivative of $\frac{x^{5/3}}{5/3}$
♦ 1, is the derivative of x
2. So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[x^{2/3} \,+\,1 \right] \, dx}} & {~=~} &{\int{\left[x^{2/3} \right]\,dx}~+~\int{\left[1 \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left[\frac{x^{5/3}}{5/3}\,+\,C_1 \right] ~+~\left[x\,+\,C_2 \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{3 x^{5/3}}{5}\,+\, x \,+\,C_1 \,+\,C_2} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{3 x^{5/3}}{5}\,+\, x \,+\,C} \\
\end{array}$
•
This is the general form of the required anti derivative.
◼ Remarks:
4 (magenta color): Addition/subtraction of any two real numbers C1 and C2 will give another real number which can be denoted in general as C.
• Or we can write:
$\int{\left[x^{2/3} \,+\,1 \right] \, dx}~=~\frac{3 x^{5/3}}{5}\,+\, x$ is an anti
derivative of $x^{2/3} \,+\,1$.
Part (iii):
1. We want $\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x^{3/2} \right]\,dx}~+~\int{\left[2 e^x \right]}~-~\int{\left[\frac{1}{x} \right] \, dx}$
• We search for F and find that:
♦ x3/2, is the derivative of $\frac{x^{5/2}}{5/2}$
♦ 2ex, is the derivative of 2ex
♦ 1/x, is the derivative of log|x|
2. So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}} & {~=~} &{\int{\left[x^{3/2} \right] \, dx}~+~\int{\left[2 e^x \right] \, dx}~-~\int{\left[\frac{1}{x} \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left[\frac{x^{5/2}}{5/2}\,+\,C_1 \right]~+~\left[2 e^x \,+\,C_2 \right]~-~\left[\log|x| \,+\, C_3\right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x| \,+\,C_1 \,+\,C_2 \,-\,C_3} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x| \,+\,C} \\
\end{array}$
•
This is the general form of the required anti derivative.
◼ Remarks:
3 (magenta color): Addition/subtraction of any two real numbers C1 and C2 will give another real number which can be denoted in general as C.
• Or we can write:
$\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}~=~\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x|$ is an anti
derivative of $x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x}$.
Once we understand the basics, there is no need to write all the detailed steps. So from now on wards, we shall write only the necessary steps.
Solved example 23.3
Find the following integrals:
$(i)~\int{\left[\sin x \,+\, \cos x \right] \, dx}~~~~~~~(ii)~\int {\left[\csc x (\csc x \,+\, \cot x)\right]dx}~~~~~~~(iii)~\int{\left[\frac{1 \,-\, \sin x}{\cos^2 x} \right] \, dx}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\sin x \,+\, \cos x \right] \, dx}} & {~=~} &{\int{\left[\sin x \right] \, dx}~+~\int{\left[\cos x \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left[- \cos x\,+\,C_1 \right]~+~\left[\sin x \,+\,C_2 \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{-\cos x \,+\, \sin x \,+\, C} \\
\end{array}$
Part (ii):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\csc x (\csc x \,+\, \cot x) \right] \, dx}} & {~=~} &{\int{\left[\csc^2 x \,+\, \csc x \cot x \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\int{\left[\csc^2 x \right] \, dx}~+~\int{\left[\csc x \cot x \right] \, dx}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left[-\cot x \,+\, C_1 \right]~+~\left[-\csc x \,+\,C_2 \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{-\cot x \,-\, \csc x \,+\,C} \\
\end{array}$
Part (iii):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1 \,-\, \sin x}{\cos^2 x} \right] \, dx}} & {~=~} &{\int{\left[\frac{1 }{\cos^2 x} ~-~\frac{\sin x}{\cos^2 x}\right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\int{\left[\sec^2 x ~-~ \sec x \tan x \right] \, dx}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\int{\left[\sec^2 x \right] \, dx}~-~\int{\left[\sec x \tan x \right] \, dx}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\left[\tan x \,+\, C_1 \right]~-~\left[\sec x \,+\,C_2 \right]} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\tan x \,-\, \sec x \,+\,C} \\
\end{array}$
Part (iv):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{4}{1+x^2} \right] \, dx}} & {~=~} &{4 \int{\left[\frac{1}{1+x^2} \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{4 \left[\tan^{-1} x \,+\, C \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{4 \tan^{-1} x \,+\, C} \\
\end{array}$
Solved example 23.4
Find the anti derivative F of f defined by f (x) = 4x3 − 6, where F(0) = 3
Solution:
1. First we will write the general form of the anti derivative:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[4x^3 \,-\,6 \right] \, dx}} & {~=~} &{\int{\left[4 x^3 \right] \, dx}~-~\int{\left[6 \right] \, dx}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left[\frac{4 x^4}{4} \,+\, C_1 \right]~-~\left[6x \,+\,C_2 \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{x^4 \,-\, 6x \,+\,C} \\
\end{array}$
•
So the general form of the anti derivative is: x4 − 6x + C
2.Given that, when x = 0, the anti derivative becomes 3.
•
So we can write: x4 − 6x + C = 3
⇒ (0)4 − 6(0) + C = 3
⇒ C = 3
3. So based on the general form, we can write:
The exact anti derivative is: x4 − 6x + 3
The link below gives a few more solved examples:
In the next section, we will see Methods of Integration.
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