
How do you integrate \[\int{\dfrac{1}{{{x}^{\dfrac{3}{2}}}+{{x}^{\dfrac{1}{2}}}}dx}\]?
Answer
543.3k+ views
Hint: In the given question, we have been asked to integrate the following function. In order to solve the question, we need to use the formulas and the concepts of integration. Here first we transform the integral by using the law of exponent i.e. \[\dfrac{1}{{{x}^{a}}}={{x}^{-a}}\]. Later we simplify the integral and use substitution method for further integration and we will get our required answer.
Complete step by step solution:
We have given,
\[\Rightarrow \int{\dfrac{1}{{{x}^{\dfrac{3}{2}}}+{{x}^{\dfrac{1}{2}}}}dx}\]
Let the given integral be I, such that,
\[\Rightarrow I=\int{\dfrac{1}{{{x}^{\dfrac{3}{2}}}+{{x}^{\dfrac{1}{2}}}}dx}\]
Taking the common factor from the denominator, we get
\[\Rightarrow I=\int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}\left( x+1 \right)}dx}\]
As we know that \[\dfrac{1}{{{x}^{a}}}={{x}^{-a}}\]
Therefore,
\[\Rightarrow I=\int{\dfrac{{{x}^{-\ \dfrac{1}{2}}}}{\left( x+1 \right)}dx}\]
We can also write \[x={{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}\]
Now, putting this value in the denominator, we get
\[\Rightarrow I=\int{\dfrac{{{x}^{-\ \dfrac{1}{2}}}}{{{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}+1}dx}\]
\[\Rightarrow I=2\int{\dfrac{\dfrac{1}{2}{{x}^{-\ \dfrac{1}{2}}}}{{{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}+1}dx}\]
Now, substituting\[u=\left( {{x}^{\dfrac{1}{2}}} \right)\], then\[du=\dfrac{1}{2}{{x}^{-\ \dfrac{1}{2}}}dx\]
Therefore,
\[\Rightarrow I=2\int{\dfrac{1}{{{u}^{2}}+1}dx}\]
The resultant integral is the arctangent integral;
Hence,
\[I=2\arctan \left( u \right)+C\]
Undo the substitution i.e.\[u=\left( {{x}^{\dfrac{1}{2}}} \right)\], we get
\[I=2\arctan \left( {{x}^{\dfrac{1}{2}}} \right)+C\]
Hence, it is the required integration.
Note: Here, we need to remember that we have to put the constant term C after the integration equation and the value of the given constant i.e. C, it can be any value 0 equal to zero also. In order to solve the question that is given above, students need to know the basic formula of integration and they should very well keep all the standard integral into their mind because sometimes the given integration is the standard integral and we do not need to solve the question further and directly write the resultant integral.
Complete step by step solution:
We have given,
\[\Rightarrow \int{\dfrac{1}{{{x}^{\dfrac{3}{2}}}+{{x}^{\dfrac{1}{2}}}}dx}\]
Let the given integral be I, such that,
\[\Rightarrow I=\int{\dfrac{1}{{{x}^{\dfrac{3}{2}}}+{{x}^{\dfrac{1}{2}}}}dx}\]
Taking the common factor from the denominator, we get
\[\Rightarrow I=\int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}\left( x+1 \right)}dx}\]
As we know that \[\dfrac{1}{{{x}^{a}}}={{x}^{-a}}\]
Therefore,
\[\Rightarrow I=\int{\dfrac{{{x}^{-\ \dfrac{1}{2}}}}{\left( x+1 \right)}dx}\]
We can also write \[x={{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}\]
Now, putting this value in the denominator, we get
\[\Rightarrow I=\int{\dfrac{{{x}^{-\ \dfrac{1}{2}}}}{{{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}+1}dx}\]
\[\Rightarrow I=2\int{\dfrac{\dfrac{1}{2}{{x}^{-\ \dfrac{1}{2}}}}{{{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}+1}dx}\]
Now, substituting\[u=\left( {{x}^{\dfrac{1}{2}}} \right)\], then\[du=\dfrac{1}{2}{{x}^{-\ \dfrac{1}{2}}}dx\]
Therefore,
\[\Rightarrow I=2\int{\dfrac{1}{{{u}^{2}}+1}dx}\]
The resultant integral is the arctangent integral;
Hence,
\[I=2\arctan \left( u \right)+C\]
Undo the substitution i.e.\[u=\left( {{x}^{\dfrac{1}{2}}} \right)\], we get
\[I=2\arctan \left( {{x}^{\dfrac{1}{2}}} \right)+C\]
Hence, it is the required integration.
Note: Here, we need to remember that we have to put the constant term C after the integration equation and the value of the given constant i.e. C, it can be any value 0 equal to zero also. In order to solve the question that is given above, students need to know the basic formula of integration and they should very well keep all the standard integral into their mind because sometimes the given integration is the standard integral and we do not need to solve the question further and directly write the resultant integral.
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