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Question: How do you find the integral of \[f\left( x \right)={{x}^{4}}{{e}^{x}}\] by using integration by par...

How do you find the integral of f(x)=x4exf\left( x \right)={{x}^{4}}{{e}^{x}} by using integration by parts?

Explanation

Solution

In the given question, we have been asked to integrate the given expression using partial fraction. Firstly use the partial fraction method to integrate the function by using the by-parts formula i.e. ∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\int{f\left( x \right)g'\left( x \right)dx=f\left( x \right)g\left( x \right)-\int{f'\left( x \right)g\left( x \right)dx}}. We will continue the process of applying the integration by-parts till we get the final solution.

Complete step by step answer:
We have, ∫x4exdx\int{{{x}^{4}}{{e}^{x}}dx}. In order to integrate ∫x4exdx\int{{{x}^{4}}{{e}^{x}}dx} using partial fractions, we will first identify two factors in the integrand. Formula of integration by parts as follows;
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\int{f\left( x \right)g'\left( x \right)dx=f\left( x \right)g\left( x \right)-\int{f'\left( x \right)g\left( x \right)dx}}
Thus, integrating the resultant expression, we obtain
Here,
Let f(x)=x4 then f′(x)=4x3f\left( x \right)={{x}^{4}}\ then\ f'\left( x \right)=4{{x}^{3}}
And g′(x)=ex then g(x)=exg'\left( x \right)={{e}^{x}}\ then\ g\left( x \right)={{e}^{x}}
Therefore,
⇒∫x4exdx=x4ex−∫4x3exdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-\int{4{{x}^{3}}{{e}^{x}}dx}
Taking the constant part out of the integration, we have
⇒∫x4exdx=x4ex−4∫x3exdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4\int{{{x}^{3}}{{e}^{x}}dx}

Now similarly,
Integrating further using by-parts; we obtained
⇒∫x4exdx=x4ex−4x3ex−4∫3x2exdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4{{x}^{3}}{{e}^{x}}-4\int{3{{x}^{2}}{{e}^{x}}dx}
Again taking the constant part out of the integration, we have
⇒∫x4exdx=x4ex−4xex−12∫x2exdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4x{{e}^{x}}-12\int{{{x}^{2}}{{e}^{x}}dx}
Now similarly,
Integrating further using by-parts; we obtained
⇒∫x4exdx=x4ex−4xex−12x2ex−12∫2xexdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4x{{e}^{x}}-12{{x}^{2}}{{e}^{x}}-12\int{2x{{e}^{x}}dx}
Again taking the constant part out of the integration, we have
⇒∫x4exdx=x4ex−4xex−12x2ex−24∫xexdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4x{{e}^{x}}-12{{x}^{2}}{{e}^{x}}-24\int{x{{e}^{x}}dx}

Now similarly,
Integrating further using by-parts; we obtained
⇒∫x4exdx=x4ex−4xex−12x2ex−24xex−24∫1exdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4x{{e}^{x}}-12{{x}^{2}}{{e}^{x}}-24x{{e}^{x}}-24\int{1{{e}^{x}}dx}
Again taking the constant part out of the integration, we have
⇒∫x4exdx=x4ex−4xex−12x2ex−24xex−24∫exdx\Rightarrow \int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4x{{e}^{x}}-12{{x}^{2}}{{e}^{x}}-24x{{e}^{x}}-24\int{{{e}^{x}}dx}
As we know that,
∫exdx=ex\int{{{e}^{x}}dx={{e}^{x}}}
⇒∫x4exdx=x4ex−4xex−12x2ex−24xex−24ex+C\Rightarrow\int{{{x}^{4}}{{e}^{x}}dx}={{x}^{4}}{{e}^{x}}-4x{{e}^{x}}-12{{x}^{2}}{{e}^{x}}-24x{{e}^{x}}-24{{e}^{x}}+C
Taking out the common factor and simplify the above expression, we get
∴∫x4exdx=ex(x4−4x−12x2−24x−24)+C\therefore\int{{{x}^{4}}{{e}^{x}}dx}={{e}^{x}}\left( {{x}^{4}}-4x-12{{x}^{2}}-24x-24 \right)+C

Hence,the integral of f(x)=x4exf\left( x \right)={{x}^{4}}{{e}^{x}} by using integration by parts is ex(x4−4x−12x2−24x−24)+C{{e}^{x}}\left( {{x}^{4}}-4x-12{{x}^{2}}-24x-24 \right)+C.

Note: Students need to remember the concept of integration by using the by-parts method. When doing indefinite integration, always write the +C part after the integration. This +C part indicates the constant part remains after integration and can be understood when you explore it graphically. The finite integration constant gets cancelled out, so we only write it in indefinite integration.