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Question

Question: How do you integrate \(\int {{x^2}{e^{2x}}} \) by parts?...

How do you integrate x2e2x\int {{x^2}{e^{2x}}} by parts?

Explanation

Solution

In this question we have the integration of two terms in multiplication therefore; we will use the formula of integration by parts and simplify it to get the final answer.

Formula used: uvdx=uvdx(dudxvdx)\int {uvdx} = u\int v dx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)}

Complete step-by-step solution:
We have the expression as:
x2e2xdx\Rightarrow \int {{x^2}{e^{2x}}} dx
Now it has 22 terms which are x2{x^2} and e2x{e^{2x}}, in this question we will consider the first part to be uu and the latter part as vv and apply the formula of integration by parts.
Now on using the formula for integration by parts on the expression, we get:
x2e2xdx=x2e2xdx(d(x2)dxe2xdx)\Rightarrow \int {{x^2}{e^{2x}}dx} = {x^2}\int {{e^{2x}}} dx - \int {\left( {\dfrac{{d({x^2})}}{{dx}}\int {{e^{2x}}dx} } \right)}
Now on integrating the first term, we get:
x2e2xdx=x2×e2x2(d(x2)dxe2xdx)\Rightarrow \int {{x^2}{e^{2x}}dx} = {x^2} \times \dfrac{{{e^{2x}}}}{2} - \int {\left( {\dfrac{{d({x^2})}}{{dx}}\int {{e^{2x}}dx} } \right)}
Now on simplifying the first term, we get:
x2e2xdx=x2e2x2(d(x2)dxe2xdx)\Rightarrow \int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \int {\left( {\dfrac{{d({x^2})}}{{dx}}\int {{e^{2x}}dx} } \right)}
Now the first part is solved therefore, we will now proceed to the latter part of the expression. On taking the derivative, we get:
x2e2xdx=x2e2x2(2xe2xdx)\Rightarrow \int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \int {\left( {2x\int {{e^{2x}}dx} } \right)}
Now on taking the integration, we get:
x2e2xdx=x2e2x2(2x×e2x2)\Rightarrow \int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \int {\left( {2x \times \dfrac{{{e^{2x}}}}{2}} \right)}
Now on cancelling the terms and simplifying the expression, we get:
x2e2xdx=x2e2x2(xe2x2)\Rightarrow \int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \int {\left( {\dfrac{{x{e^{2x}}}}{2}} \right)}
Now we can take the constant value out of the integration and rewrite the expression as:
x2e2xdx=x2e2x212xe2x(1)\Rightarrow \int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \dfrac{1}{2}\int {x{e^{2x}}} \to (1)
Now the latter part cannot be integrated directly therefore, we will use the integration by parts formula on it by considering uu as xx and vv as e2x{e^{2x}}.
We will solve the integration separately and substitute it in the equation. On using the formula on xe2x\int {x{e^{2x}}} we get:
xe2xdx=xe2xdx(d(x)dxe2xdx)\Rightarrow \int {x{e^{2x}}dx} = x\int {{e^{2x}}} dx - \int {\left( {\dfrac{{d(x)}}{{dx}}\int {{e^{2x}}dx} } \right)}
On simplifying, we get:
xe2xdx=x×e2x2(1×e2x2)\Rightarrow \int {x{e^{2x}}dx} = x \times \dfrac{{{e^{2x}}}}{2} - \int {\left( {1 \times \dfrac{{{e^{2x}}}}{2}} \right)}
On simplifying, we get:
xe2xdx=xe2x212e2x\Rightarrow \int {x{e^{2x}}dx} = \dfrac{{x{e^{2x}}}}{2} - \dfrac{1}{2}\int {{e^{2x}}}
On integrating the term, we get:
xe2xdx=xe2x212×e2x2\Rightarrow \int {x{e^{2x}}dx} = \dfrac{{x{e^{2x}}}}{2} - \dfrac{1}{2} \times \dfrac{{{e^{2x}}}}{2}
On simplifying, we get:
xe2xdx=xe2x2e2x4\Rightarrow \int {x{e^{2x}}dx} = \dfrac{{x{e^{2x}}}}{2} - \dfrac{{{e^{2x}}}}{4}
Now on substituting it back in equation (1)(1) we get:
x2e2xdx=x2e2x2xe2x4e2x8\Rightarrow \int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \dfrac{{x{e^{2x}}}}{4} - \dfrac{{{e^{2x}}}}{8}

x2e2xdx=x2e2x2xe2x4e2x8\int {{x^2}{e^{2x}}dx} = \dfrac{{{x^2}{e^{2x}}}}{2} - \dfrac{{x{e^{2x}}}}{4} - \dfrac{{{e^{2x}}}}{8} is the required solution.

Note: It is to be remembered that while doing integration by parts the terms uu and vv should follow the sequence of the acronym LIATELIATE, which stands for logarithm, inverse, algebraic, trigonometric and exponential respectively.
It is to be remembered that integration and differentiation are inverse of each other. If the integration of aa is bb then the derivative of bb is aa.