Question

How do you integrate \[\int{\dfrac{{{x}^{3}}{{e}^{{{x}^{2}}}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx}\]?

Answer 1

Hint: This question is from the topic of integration. In solving this question, we will first do the substitution. We will substitute the term \[{{x}^{2}}\] by t. After that, we will do the further process. After that, we will do the integration by parts using the ILATE rule. After solving the further question, we will get our answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to integrate \[\int{\dfrac{{{x}^{3}}{{e}^{{{x}^{2}}}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx}\].
Let integration be
\[I=\int{\dfrac{{{x}^{3}}{{e}^{{{x}^{2}}}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx}\]
The above integration can also be written as
\[I=\int{\dfrac{{{x}^{2}}{{e}^{{{x}^{2}}}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot xdx}\]
Now, we will use the substitution method to solve the further integration.
Substitute
\[{{x}^{2}}=t\]
Now, differentiate the above equation, we will get
\[2x\cdot dx=dt\]
The above equation can also be written as
\[\Rightarrow x\cdot dx=\dfrac{1}{2}dt\]
Now, putting \[{{x}^{2}}=t\] and \[x\cdot dx=\dfrac{1}{2}dt\] in the integration \[I=\int{\dfrac{{{x}^{2}}{{e}^{{{x}^{2}}}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot xdx}\], we will get
\[I=\int{\dfrac{t\cdot {{e}^{t}}}{{{\left( t+1 \right)}^{2}}}\cdot \dfrac{1}{2}dt}\]
The above integration can also be written as
\[\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( t+1-1 \right){{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}\]
The above integration can also be written as
\[\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( t+1 \right){{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}-\dfrac{1}{2}\int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}\]
We can write the above integration as
\[\Rightarrow 2I=\left[ \int{\dfrac{\left( t+1 \right){{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]-\left[ \int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]\]
The above can also be written as
\[\Rightarrow 2I=\left[ \int{\dfrac{{{e}^{t}}}{\left( t+1 \right)}dt} \right]-\left[ \int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]\]
Now, we will do integration by parts for further differentiation. We will use the ILATE rule here. The ILATE rule says that whenever we have to integrate the two functions. Then, the integration will be like:
\[\int{f\left( t \right)g\left( t \right)dt}=f\left( t \right)\int{g\left( t \right)dt}-\int{\left[ \int{g\left( t \right)dt} \right]\dfrac{d}{dt}f\left( t \right)dt}\], where f(t) is the first function and g(t) is the second function. We will take f(t) as the first function according to the order of ILATE (that is Inverse function, Logarithmic function, Algebraic function, Trigonometric function, and Exponential function).
So, in the integration \[\Rightarrow 2I=\left[ \int{\dfrac{{{e}^{t}}}{\left( t+1 \right)}dt} \right]-\left[ \int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]\], we will take \[{{e}^{t}}\] as second function because it is the exponential function and according to ILATE, it comes in the last function.
We can write the above integration as
\[\Rightarrow 2I=\left[ \dfrac{1}{\left( t+1 \right)}\int{{{e}^{t}}dt}-\int{\left[ \int{{{e}^{t}}dt} \right]\dfrac{d}{dt}\dfrac{1}{\left( t+1 \right)}dt} \right]-\left[ \int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]\]
Using the formula \[\int{{{e}^{t}}dt}={{e}^{t}}\], we can write the above as
\[\Rightarrow 2I=\left[ \dfrac{1}{\left( t+1 \right)}{{e}^{t}}-\int{{{e}^{t}}\dfrac{d}{dt}\dfrac{1}{\left( t+1 \right)}dt} \right]-\left[ \int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]\]
Now, using the formula \[\dfrac{d}{dt}\dfrac{1}{\left( t+1 \right)}=-\dfrac{1}{{{\left( t+1 \right)}^{2}}}\], we can write the above integration as
\[\Rightarrow 2I=\left[ \dfrac{1}{\left( t+1 \right)}{{e}^{t}}-\int{{{e}^{t}}\left\{ -\dfrac{1}{{{\left( t+1 \right)}^{2}}} \right\}dt} \right]-\left[ \int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt} \right]\]
The above integration can also be written as
\[\Rightarrow 2I=\dfrac{1}{\left( t+1 \right)}{{e}^{t}}-\int{{{e}^{t}}\left\{ -\dfrac{1}{{{\left( t+1 \right)}^{2}}} \right\}dt}-\int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}\]
\[\Rightarrow 2I=\dfrac{1}{\left( t+1 \right)}{{e}^{t}}-\int{\left\{ -\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}} \right\}dt}-\int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}\]
\[\Rightarrow 2I=\dfrac{1}{\left( t+1 \right)}{{e}^{t}}+\int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}-\int{\dfrac{{{e}^{t}}}{{{\left( t+1 \right)}^{2}}}dt}\]
The above can also be written as
\[\Rightarrow 2I=\dfrac{1}{\left( t+1 \right)}{{e}^{t}}\]
\[\Rightarrow I=\dfrac{1}{2}\dfrac{{{e}^{t}}}{\left( t+1 \right)}\]
As we know that after integration, a constant is always added, so we can write the equation as
\[\Rightarrow I=\dfrac{1}{2}\dfrac{{{e}^{t}}}{\left( t+1 \right)}+C\], where C is any constant
Now, putting the value of t as square of x as we have taken above, we can write
\[\Rightarrow I=\dfrac{1}{2}\dfrac{{{e}^{{{x}^{2}}}}}{\left( {{x}^{2}}+1 \right)}+C\]
Hence, we got that the integration of \[\int{\dfrac{{{x}^{2}}{{e}^{{{x}^{2}}}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot xdx}\] is
\[\dfrac{1}{2}\dfrac{{{e}^{{{x}^{2}}}}}{\left( {{x}^{2}}+1 \right)}+C\]

Note: We should have a better knowledge in the topic of integration to solve this type of question easily. We should know about the ILATE rule. We should know how to do integration by parts. The integration by parts formula is:
\[\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left[ \int{g\left( x \right)dx} \right]\dfrac{d}{dx}f\left( x \right)dx}\], where f(x) and g(x) are two functions of x. We have taken f(x) as the first function and g(x) as the second function. We will take the first function and second function according to the order of ILATE (that is Inverse function, Logarithmic function, Algebraic function, Trigonometric function, and Exponential function)
Remember the following formula:
\[\int{{{e}^{t}}dt}={{e}^{t}}\]
\[\dfrac{d}{dx}\dfrac{1}{\left( 1+x \right)}=-\dfrac{1}{{{\left( 1+x \right)}^{2}}}\]