What is the integral of $\int{\dfrac{{{x}^{3}}}{1+{{x}^{2}}}dx}$?
Answer
589.2k+ views
Hint: To obtain the integral of the given integrand we will find what type of function it is. Firstly we will check the degree of polynomial of numerator and denominator to know what type of function it is. Then we will make some changes in the numerator so that we get a value that can be integrated easily. Finally by using integration techniques we will get the desired answer.
Complete step-by-step solution:
We have to find the integral of:
$\int{\dfrac{{{x}^{3}}}{1+{{x}^{2}}}dx}$
As we can see that the degree of polynomial in numerator is 3 and degree of polynomial in denominator is 2 so,
$3>2$
So it is an Improper Rational Function.
Now, we will proceed forward by making the change in numerator as follows:
$\begin{align}
& \Rightarrow \int{\dfrac{{{x}^{3}}+x-x}{1+{{x}^{2}}}dx} \\
& \Rightarrow \int{\dfrac{x\left( {{x}^{2}}+1 \right)-x}{1+{{x}^{2}}}dx} \\
\end{align}$
Next separate the two terms in the numerator as follows:
$\Rightarrow \int{\left( \dfrac{x\left( {{x}^{2}}+1 \right)}{1+{{x}^{2}}}-\dfrac{x}{1+{{x}^{2}}} \right)dx}$
$\Rightarrow \int{\left( x-\dfrac{x}{1+{{x}^{2}}} \right)dx}$…..$\left( 1 \right)$
So we will use the below formula for first and second term in above equation:
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$…..$\left( 2 \right)$
$\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx=\ln f\left( x \right)}$….$\left( 3 \right)$
For using formula (2) in equation (1) we will multiply and divide it by 2 as follows:
$\Rightarrow \int{xdx}-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}dx}$
By using formula (2) and (3) in above value we get,
$\Rightarrow \dfrac{{{x}^{2}}}{2}-\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C$
Hence integral of $\int{\dfrac{{{x}^{3}}}{1+{{x}^{2}}}dx}$ is $\dfrac{{{x}^{2}}}{2}-\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C$ where $C$ is any constant.
Note: The process by which we find the integrals is known as integration. Integration is a very important operation of calculus. It assigns numbers to function in such a way that describes area, displacement volume and many other concepts. Integration is opposite of differentiation and the sign used to represent integration is $\int{{}}$. Different methods of integrating a function may give different answers but they only differ by a constant.
Complete step-by-step solution:
We have to find the integral of:
$\int{\dfrac{{{x}^{3}}}{1+{{x}^{2}}}dx}$
As we can see that the degree of polynomial in numerator is 3 and degree of polynomial in denominator is 2 so,
$3>2$
So it is an Improper Rational Function.
Now, we will proceed forward by making the change in numerator as follows:
$\begin{align}
& \Rightarrow \int{\dfrac{{{x}^{3}}+x-x}{1+{{x}^{2}}}dx} \\
& \Rightarrow \int{\dfrac{x\left( {{x}^{2}}+1 \right)-x}{1+{{x}^{2}}}dx} \\
\end{align}$
Next separate the two terms in the numerator as follows:
$\Rightarrow \int{\left( \dfrac{x\left( {{x}^{2}}+1 \right)}{1+{{x}^{2}}}-\dfrac{x}{1+{{x}^{2}}} \right)dx}$
$\Rightarrow \int{\left( x-\dfrac{x}{1+{{x}^{2}}} \right)dx}$…..$\left( 1 \right)$
So we will use the below formula for first and second term in above equation:
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$…..$\left( 2 \right)$
$\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx=\ln f\left( x \right)}$….$\left( 3 \right)$
For using formula (2) in equation (1) we will multiply and divide it by 2 as follows:
$\Rightarrow \int{xdx}-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}dx}$
By using formula (2) and (3) in above value we get,
$\Rightarrow \dfrac{{{x}^{2}}}{2}-\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C$
Hence integral of $\int{\dfrac{{{x}^{3}}}{1+{{x}^{2}}}dx}$ is $\dfrac{{{x}^{2}}}{2}-\dfrac{1}{2}\ln \left( 1+{{x}^{2}} \right)+C$ where $C$ is any constant.
Note: The process by which we find the integrals is known as integration. Integration is a very important operation of calculus. It assigns numbers to function in such a way that describes area, displacement volume and many other concepts. Integration is opposite of differentiation and the sign used to represent integration is $\int{{}}$. Different methods of integrating a function may give different answers but they only differ by a constant.
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