How do you evaluate the integral $\int{\dfrac{dx}{{{x}^{4}}-16}}$ ?
Answer
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Hint: For evaluating the integral of the given question, $\int{\dfrac{dx}{{{x}^{4}}-16}}$ , we will expand the equation so that it becomes easy to solve. Then, we will apply integration in the expression of the question and will use the required formula to get the value of the given question. After that we will simplify the integrated value if possible. Thus we will get the integral of the question.
Complete step by step solution:
Since, we have the question as:
$\Rightarrow \int{\dfrac{dx}{{{x}^{4}}-16}}$ … $\left( i \right)$
Now, we will write the above equation without integration so that we can expand it as:
\[\Rightarrow \dfrac{1}{{{x}^{4}}-16}\]
As we know that $16$ is square of $4$ , so we can write above equation as:
\[\Rightarrow \dfrac{1}{{{x}^{4}}-{{4}^{2}}}\] … $\left( ii \right)$
Here, we will use the formula $\left( {{a}^{2}}-{{b}^{2}} \right)$ that is:
$\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
Similarly:
$\Rightarrow \left( {{x}^{2}}-{{4}^{2}} \right)=\left( x+4 \right)\left( x-4 \right)$ … $\left( iii \right)$
So, we can use the above method in the equation $\left( ii \right)$ that will be as:
\[\Rightarrow \dfrac{1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-4 \right)}\]
Now, we will expand this equation in two terms as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]\] … $\left( iv \right)$
Here we will expand the term \[\dfrac{1}{\left( {{x}^{2}}-4 \right)}\] as:
\[\Rightarrow \dfrac{1}{\left( {{x}^{2}}-4 \right)}\]
Since, $4$ is a square of $2$ . So we can write it as:
\[\Rightarrow \dfrac{1}{\left( {{x}^{2}}-{{2}^{2}} \right)}\]
By using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, we can convert the above equation as:
\[\Rightarrow \dfrac{1}{\left( x-2 \right)\left( x+2 \right)}\]
Now, we can expand the above equation as:
\[\Rightarrow \dfrac{1}{4}\left[ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right]\]
Since, we got the expansion of the term \[\dfrac{1}{{{x}^{2}}-4}\] as \[\dfrac{1}{4}\left[ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right]\] . So, we will use this equation in equation $\left( iv \right)$ so that we can able to solve the question as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\left\{ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right\}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]\]
Now, we will use the above equation as a replacement in equation $\left( i \right)$ as:
$\Rightarrow \int{\dfrac{dx}{{{x}^{4}}-16}}=\int{\dfrac{1}{8}\left[ \dfrac{1}{4}\left\{ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right\}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx$
After opening the middle bracket, we will have the above equation as:
$\Rightarrow \int{\dfrac{dx}{{{x}^{4}}-16}}$
$\Rightarrow \int{\dfrac{1}{8}\left[ \dfrac{1}{4}\times \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{4}\times \dfrac{1}{\left( x+2 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx$
Now, we will apply the integration in every term of the above equation as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\int{\dfrac{1}{\left( x-2 \right)}}dx-\dfrac{1}{4}\int{\dfrac{1}{\left( x+2 \right)}dx}-\int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx} \right]\]
Since, $4$ is a square of $2$ . So we can write the above equation as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\int{\dfrac{1}{\left( x-2 \right)}}dx-\dfrac{1}{4}\int{\dfrac{1}{\left( x+2 \right)}dx}-\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx} \right]\] … $\left( v \right)$
When a term is in the form of $\dfrac{1}{x-a}$ , the integration this term is $\log \left( x-a \right)$ . Similarly, the integration of $\dfrac{1}{x-a}$ is $\log \left( x-a \right)$ . So, we have the integration of these terms as:
\[\Rightarrow \int{\dfrac{1}{\left( x+2 \right)}}dx=\log \left( x+2 \right)+{{c}_{1}}\] … $\left( 1 \right)$
And
\[\Rightarrow \int{\dfrac{1}{\left( x-2 \right)}}dx=\log \left( x-2 \right)+{{c}_{2}}\] … $\left( 2 \right)$
Here, we will evaluate \[\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}\] as:
Let us consider that
$\Rightarrow x=2\tan \theta $
$\Rightarrow \tan \theta =\dfrac{x}{2}$
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{x}{2} \right)$
Now, we will differentiate it with respect to $\theta $ , we will have:
$\Rightarrow dx=2{{\sec }^{2}}\theta \text{ }d\theta $
Now, we will put these values in the question that is \[\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}\] as:
\[\Rightarrow \int{\dfrac{1}{\left( {{\left[ 2\tan \theta \right]}^{2}}+{{2}^{2}} \right)}{{\sec }^{2}}\theta \text{ }d\theta }\]
Now, we will solve the above equation as:
\[\Rightarrow \int{\dfrac{1}{\left( 4{{\tan }^{2}}\theta +4 \right)}{{\sec }^{2}}\theta \text{ }d\theta }\]
\[\Rightarrow \int{\dfrac{1}{4\left( {{\tan }^{2}}\theta +1 \right)}{{\sec }^{2}}\theta \text{ }d\theta }\]
Since, we know that ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ . So the above equation will be as:
\[\Rightarrow \int{\dfrac{1}{4{{\sec }^{2}}\theta }{{\sec }^{2}}\theta \text{ }d\theta }\]
Now, ${{\sec }^{2}}\theta $ will be eliminated from the above equation as:
\[\Rightarrow \int{\dfrac{1}{4}\text{ }d\theta }\]
Now, we have the integrated value as:
\[\Rightarrow \dfrac{1}{4}\theta +{{c}_{3}}\]
Here, we will put $\theta ={{\tan }^{-1}}\left( \dfrac{x}{2} \right)$ . So the above equation will be:
\[\Rightarrow \dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{c}_{3}}\] … $\left( 3 \right)$
Now, we will use the equation $\left( 1 \right)$, $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( v \right)$ as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\log \left( x-2 \right)+{{c}_{1}}-\dfrac{1}{4}\log \left( x+2 \right)+{{c}_{2}}-\dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{c}_{3}} \right]+{{c}_{4}}\]
Where, ${{c}_{1}}$ , ${{c}_{2}}$ , ${{c}_{3}}$ and ${{c}_{4}}$ are constants.
As we know that the subtraction rule of log:
$\Rightarrow \log a-\log b=\log \left( \dfrac{a}{b} \right)$
So, the above equation will be:
Now, we will use this method in the above expression of the question as:
$\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\log \dfrac{\left( x-a \right)}{\left( x+a \right)}-\dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+\dfrac{1}{8}\left( {{c}_{1}}+{{c}_{2}}+{{c}_{3}} \right)+{{c}_{4}}$
$\Rightarrow \dfrac{1}{8}\times \dfrac{1}{4}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)}-{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C$ Where, $C$ is also a constant.
$\Rightarrow \dfrac{1}{32}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)}-{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C$
Hence, the value of $\int{\dfrac{dx}{{{x}^{4}}-16}}$ is equal to $\dfrac{1}{32}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)}-{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C$
Note: There are some basic rules and formulas for integration that we need to remember by learning because it is very useful for solving any integral equation. From example: in the given question we have $\int{\dfrac{dx}{{{x}^{4}}-16}}$ and after expanding it we got the equation as:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx\]
The above equation can be written as:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-{{2}^{2}} \right)}-\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)} \right]}dx\]
By using the formula or rule of some particular function, we already have the integrated value of the above equation as:
$\Rightarrow \int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{1}{2a}\log \left( \dfrac{x+a}{x-a} \right)}+{{c}_{1}}$
And
$\Rightarrow \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{2a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}+{{c}_{2}}$
Now, we can use these formulas as:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx=\dfrac{1}{8}\left[ \dfrac{1}{4}\log \left( \dfrac{x+2}{x-2} \right)+{{c}_{1}}-\dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{c}_{2}} \right]\]
And after simplification we will get:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx=\dfrac{1}{32}\left[ \log \left( \dfrac{x+2}{x-2} \right)+{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C\]
Hence, the solution is correct.
Complete step by step solution:
Since, we have the question as:
$\Rightarrow \int{\dfrac{dx}{{{x}^{4}}-16}}$ … $\left( i \right)$
Now, we will write the above equation without integration so that we can expand it as:
\[\Rightarrow \dfrac{1}{{{x}^{4}}-16}\]
As we know that $16$ is square of $4$ , so we can write above equation as:
\[\Rightarrow \dfrac{1}{{{x}^{4}}-{{4}^{2}}}\] … $\left( ii \right)$
Here, we will use the formula $\left( {{a}^{2}}-{{b}^{2}} \right)$ that is:
$\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
Similarly:
$\Rightarrow \left( {{x}^{2}}-{{4}^{2}} \right)=\left( x+4 \right)\left( x-4 \right)$ … $\left( iii \right)$
So, we can use the above method in the equation $\left( ii \right)$ that will be as:
\[\Rightarrow \dfrac{1}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-4 \right)}\]
Now, we will expand this equation in two terms as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]\] … $\left( iv \right)$
Here we will expand the term \[\dfrac{1}{\left( {{x}^{2}}-4 \right)}\] as:
\[\Rightarrow \dfrac{1}{\left( {{x}^{2}}-4 \right)}\]
Since, $4$ is a square of $2$ . So we can write it as:
\[\Rightarrow \dfrac{1}{\left( {{x}^{2}}-{{2}^{2}} \right)}\]
By using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, we can convert the above equation as:
\[\Rightarrow \dfrac{1}{\left( x-2 \right)\left( x+2 \right)}\]
Now, we can expand the above equation as:
\[\Rightarrow \dfrac{1}{4}\left[ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right]\]
Since, we got the expansion of the term \[\dfrac{1}{{{x}^{2}}-4}\] as \[\dfrac{1}{4}\left[ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right]\] . So, we will use this equation in equation $\left( iv \right)$ so that we can able to solve the question as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\left\{ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right\}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]\]
Now, we will use the above equation as a replacement in equation $\left( i \right)$ as:
$\Rightarrow \int{\dfrac{dx}{{{x}^{4}}-16}}=\int{\dfrac{1}{8}\left[ \dfrac{1}{4}\left\{ \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{\left( x+2 \right)} \right\}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx$
After opening the middle bracket, we will have the above equation as:
$\Rightarrow \int{\dfrac{dx}{{{x}^{4}}-16}}$
$\Rightarrow \int{\dfrac{1}{8}\left[ \dfrac{1}{4}\times \dfrac{1}{\left( x-2 \right)}-\dfrac{1}{4}\times \dfrac{1}{\left( x+2 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx$
Now, we will apply the integration in every term of the above equation as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\int{\dfrac{1}{\left( x-2 \right)}}dx-\dfrac{1}{4}\int{\dfrac{1}{\left( x+2 \right)}dx}-\int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx} \right]\]
Since, $4$ is a square of $2$ . So we can write the above equation as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\int{\dfrac{1}{\left( x-2 \right)}}dx-\dfrac{1}{4}\int{\dfrac{1}{\left( x+2 \right)}dx}-\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx} \right]\] … $\left( v \right)$
When a term is in the form of $\dfrac{1}{x-a}$ , the integration this term is $\log \left( x-a \right)$ . Similarly, the integration of $\dfrac{1}{x-a}$ is $\log \left( x-a \right)$ . So, we have the integration of these terms as:
\[\Rightarrow \int{\dfrac{1}{\left( x+2 \right)}}dx=\log \left( x+2 \right)+{{c}_{1}}\] … $\left( 1 \right)$
And
\[\Rightarrow \int{\dfrac{1}{\left( x-2 \right)}}dx=\log \left( x-2 \right)+{{c}_{2}}\] … $\left( 2 \right)$
Here, we will evaluate \[\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}\] as:
Let us consider that
$\Rightarrow x=2\tan \theta $
$\Rightarrow \tan \theta =\dfrac{x}{2}$
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{x}{2} \right)$
Now, we will differentiate it with respect to $\theta $ , we will have:
$\Rightarrow dx=2{{\sec }^{2}}\theta \text{ }d\theta $
Now, we will put these values in the question that is \[\int{\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)}dx}\] as:
\[\Rightarrow \int{\dfrac{1}{\left( {{\left[ 2\tan \theta \right]}^{2}}+{{2}^{2}} \right)}{{\sec }^{2}}\theta \text{ }d\theta }\]
Now, we will solve the above equation as:
\[\Rightarrow \int{\dfrac{1}{\left( 4{{\tan }^{2}}\theta +4 \right)}{{\sec }^{2}}\theta \text{ }d\theta }\]
\[\Rightarrow \int{\dfrac{1}{4\left( {{\tan }^{2}}\theta +1 \right)}{{\sec }^{2}}\theta \text{ }d\theta }\]
Since, we know that ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ . So the above equation will be as:
\[\Rightarrow \int{\dfrac{1}{4{{\sec }^{2}}\theta }{{\sec }^{2}}\theta \text{ }d\theta }\]
Now, ${{\sec }^{2}}\theta $ will be eliminated from the above equation as:
\[\Rightarrow \int{\dfrac{1}{4}\text{ }d\theta }\]
Now, we have the integrated value as:
\[\Rightarrow \dfrac{1}{4}\theta +{{c}_{3}}\]
Here, we will put $\theta ={{\tan }^{-1}}\left( \dfrac{x}{2} \right)$ . So the above equation will be:
\[\Rightarrow \dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{c}_{3}}\] … $\left( 3 \right)$
Now, we will use the equation $\left( 1 \right)$, $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( v \right)$ as:
\[\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\log \left( x-2 \right)+{{c}_{1}}-\dfrac{1}{4}\log \left( x+2 \right)+{{c}_{2}}-\dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{c}_{3}} \right]+{{c}_{4}}\]
Where, ${{c}_{1}}$ , ${{c}_{2}}$ , ${{c}_{3}}$ and ${{c}_{4}}$ are constants.
As we know that the subtraction rule of log:
$\Rightarrow \log a-\log b=\log \left( \dfrac{a}{b} \right)$
So, the above equation will be:
Now, we will use this method in the above expression of the question as:
$\Rightarrow \dfrac{1}{8}\left[ \dfrac{1}{4}\log \dfrac{\left( x-a \right)}{\left( x+a \right)}-\dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+\dfrac{1}{8}\left( {{c}_{1}}+{{c}_{2}}+{{c}_{3}} \right)+{{c}_{4}}$
$\Rightarrow \dfrac{1}{8}\times \dfrac{1}{4}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)}-{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C$ Where, $C$ is also a constant.
$\Rightarrow \dfrac{1}{32}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)}-{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C$
Hence, the value of $\int{\dfrac{dx}{{{x}^{4}}-16}}$ is equal to $\dfrac{1}{32}\left[ \log \dfrac{\left( x-a \right)}{\left( x+a \right)}-{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C$
Note: There are some basic rules and formulas for integration that we need to remember by learning because it is very useful for solving any integral equation. From example: in the given question we have $\int{\dfrac{dx}{{{x}^{4}}-16}}$ and after expanding it we got the equation as:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx\]
The above equation can be written as:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-{{2}^{2}} \right)}-\dfrac{1}{\left( {{x}^{2}}+{{2}^{2}} \right)} \right]}dx\]
By using the formula or rule of some particular function, we already have the integrated value of the above equation as:
$\Rightarrow \int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{1}{2a}\log \left( \dfrac{x+a}{x-a} \right)}+{{c}_{1}}$
And
$\Rightarrow \int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{2a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}+{{c}_{2}}$
Now, we can use these formulas as:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx=\dfrac{1}{8}\left[ \dfrac{1}{4}\log \left( \dfrac{x+2}{x-2} \right)+{{c}_{1}}-\dfrac{1}{4}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{c}_{2}} \right]\]
And after simplification we will get:
\[\Rightarrow \dfrac{1}{8}\int{\left[ \dfrac{1}{\left( {{x}^{2}}-4 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)} \right]}dx=\dfrac{1}{32}\left[ \log \left( \dfrac{x+2}{x-2} \right)+{{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]+C\]
Hence, the solution is correct.
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