Question

How do you integrate \[\int {{x^2}{e^{{x^2}}}} dx\] using integration by parts?

Answer 1

Hint: Here, we will use ILATE and choose our first and second term. Then we will use the integration by parts formula and substitute the value in it. Finally, we will solve the equation by integrating and differentiating terms in it to get the required answer.

Complete step-by-step answer:
We have to integrate\[\int {{x^2}{e^{{x^2}}}} dx\]
So by ILATE method
First Term \[ = u\left( x \right) = {x^2}\]
Second Term \[ = v\left( x \right) = {e^{{x^2}}}\]
But to integrate \[{e^{{x^2}}}\]we need a \[x\]so that we can use substitution.
\[u\left( x \right) = x\]
\[v\left( x \right) = x \times {e^{{x^2}}}\]
Now as we know formula of by parts integration is:
\[\int {\left( {u\left( x \right)v\left( x \right)} \right)dx = u\left( x \right)\int {v\left( x \right)dx - \int {\left( {\dfrac{{d\left( {u\left( x \right)} \right)}}{{dx}}\int {v\left( x \right)dx} } \right)dx} } } \]
Substituting the value in the above equation, we get,
\[\int {{x^2}{e^{{x^2}}}} dx = x\int {x \times {e^{{x^2}}}dx - \int {\left( {\dfrac{{d\left( x \right)}}{{dx}}\int {x \times {e^{{x^2}}}} dx} \right)dx} } \]….\[\left( 2 \right)\]
To integrate the term inside the integration sign we will let
\[{x^2} = t\]
Now, differentiating with respect to\[x\]we get,
\[\begin{array}{l}2x = \dfrac{{dt}}{{dx}}\\x \times dx = \dfrac{{dt}}{2}\end{array}\]
Substituting above value in equation (2) we get,
\[\int {{x^2}{e^{{x^2}}}} dx = x\int {\dfrac{{{e^t}}}{2}dt - \int {\left( {\dfrac{{d\left( x \right)}}{{dx}}\int {\dfrac{{{e^t}}}{2}} dt} \right)dt} } \]
Now integrating \[{e^t}\]and differentiating\[x\] we get,
\[\int {{x^2}{e^{{x^2}}}} dx = x \times \dfrac{{{e^t}}}{2} - \int {1 \times \dfrac{{{e^t}}}{2}dx} \]
Replacing back the value of \[t = {x^2}\]we get,
\[\begin{array}{l}\int {{x^2}{e^{{x^2}}}} dx = \dfrac{{x{e^{{x^2}}}}}{2} - \int {\dfrac{{{e^{{x^2}}}}}{2}dx} \\\int {{x^2}{e^{{x^2}}}} dx = \dfrac{{x{e^{{x^2}}}}}{2} - \dfrac{1}{2}\int {{e^{{x^2}}}dx} \end{array}\]
Now we can’t further solve this equation because \[\int {{e^{{x^2}}}dx} \] has no closed form solution.

Note:
Integration is used to calculate areas, volume, and displacement. Integration is an inverse process of differentiation. The two types of integral are definite and Indefinite integral. Definite integral has both upper and lower limits on a real line. Indefinite integral on the other side is without defined upper and lower limits. Production rule is used when we have to find the integration of two values inside the integration side; this method is also known as Integration by parts. By using integration by parts we reduce the integrals in their standard form.