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Question: What is the antiderivative of\[x{{e}^{x}}\]?...

What is the antiderivative ofxexx{{e}^{x}}?

Explanation

Solution

In this particular problem it has mentioned that the antiderivative of xexx{{e}^{x}}that means we have to integrate thexexx{{e}^{x}}by using theu.vu.v rule of integration. If the derivative is asked then we differentiate but in the question of the antiderivative we have to integrate and solve further. You will get the answer.

Complete step by step solution:
In this type of problems we have to find the antiderivative ofxexx{{e}^{x}}
It is in the form of u.vu.v rule of integration by comparing this rule with the function which is given in question that isxexx{{e}^{x}}.
By comparing this we get the value of u as well as v.
According to LIATE technique that is Log Inverse Algebra Trigonometry Exponential
u=xu=x
v=exv={{e}^{x}}
Integration ofxexx{{e}^{x}}can be written as ∫xexdx\int{x{{e}^{x}}dx}
If you observe carefully then this function in the form of u.vu.v rule
Therefore u.vu.v rule of integration can be written as
That is,
∫u.vdx=u.∫vdx−∫(dudx.∫vdx)dx\int{u.vdx=u.\int{v}}dx-\int{\left( \dfrac{du}{dx}.\int{vdx} \right)}dx
Substitute the value u and v on the above formula we get:
∫xexdx=x.∫exdx−∫(d(x)dx.∫exdx)dx\int{x{{e}^{x}}dx=x.\int{{{e}^{x}}}}dx-\int{\left( \dfrac{d(x)}{dx}.\int{{{e}^{x}}dx} \right)}dx
By using the basic rule of derivative as well as integration that is ∫exdx=ex\int{{{e}^{x}}}dx={{e}^{x}} as well as d(x)dx=1\dfrac{d(x)}{dx}=1 substitute this value in above equation we get:
∫xexdx=x.ex−∫(1.ex)dx\int{x{{e}^{x}}dx=x.{{e}^{x}}}-\int{\left( 1.{{e}^{x}} \right)}dx
By simplifying further we get:
∫xexdx=x.ex−∫exdx\int{x{{e}^{x}}dx=x.{{e}^{x}}}-\int{{{e}^{x}}}dx
Now, we have to again use the formula for substituting the formula that is ∫exdx=ex\int{{{e}^{x}}}dx={{e}^{x}} in this above equation we get:
∫xexdx=x.ex−ex+C\int{x{{e}^{x}}dx=x.{{e}^{x}}}-{{e}^{x}}+C
By taking the common ex{{e}^{x}}from this above equation we get:
∫xexdx=ex(x−1)+C\int{x{{e}^{x}}dx={{e}^{x}}(x-1)+C}
Therefore, the final answer is∫xexdx=ex(x−1)+C\int{x{{e}^{x}}dx={{e}^{x}}(x-1)+C}.

Note:
In this type of problem always remember the formula for u.vu.v rule of integration. Don’t make silly mistakes while solving and simplifying the problem. Another method to solve the u.vu.v rule of integration is Leibnitz theorem. You can also direct formulas for integration which you can see in the above problem. So, the above solution is preferred for such types of problems.