Question

How do you integrate $x{e^{{x^2}}}dx$?

Answer 1

Here we are given the term which we need to integrate and we do not have any general formula for this. So we can let the term ${x^2} = t$ and we will get $2xdx = dt{\text{ and }}xdx = \dfrac{{dt}}{2}$.
So we can substitute this value and solve it easily.

Complete step by step solution:
Here in the above problem, we are asked to integrate the term which is given as $x{e^{{x^2}}}dx$
So we need to find the answer to the problem $\int {x{e^{{x^2}}}dx}$ and we do not have any proper or direct formula for it. But we know the formula that $\int {{e^x}dx} = {e^x} + c$ and therefore we need to convert this given problem also in this form.
So we have $\int {x{e^{{x^2}}}dx}$ $- - - (1)$
So we can let ${x^2} = t$
Now if we differentiate the equation we have let, we will get:
$\dfrac{d}{{dx}}\left( {{x^2}} \right) = \dfrac{{dt}}{{dx}} \\\ \Rightarrow 2x = \dfrac{{dt}}{{dx}} \\\ \Rightarrow xdx = \dfrac{{dt}}{2} - - - - - (2) \\\$
Now we need to substitute the above value we get in equation (2) and also ${x^2} = t$ in equation (1)
$\int {{e^t}\dfrac{{dt}}{2}}$
Now we know that $\int {{e^x}dx} = {e^x} + c$
Hence we can say that:
$\int {{e^t}\dfrac{{dt}}{2}} = \dfrac{1}{2}\int {{e^t}dt}$
This is because we know that $\int {a{e^x}dx} = a\int {{e^x}dx}$
So we can take the constant outside the integration and get the value of the required integration:
Now we have got:
=$\int {{e^t}\dfrac{{dt}}{2}} = \dfrac{1}{2}\int {{e^t}dt}$
=$\int {{e^t}\dfrac{{dt}}{2}} = \dfrac{1}{2}{e^t}$
Now we have got the result in the terms of $t$ but we need it in the terms of $x$
So we can substitute now $t = {x^2}$ in it as we have let it initially for our convenience.
=$\int {x{e^{{x^2}}}dx} = \dfrac{1}{2}{e^{{x^2}}} + c$
Here $c$ is the constant of proportionality.
So in these types of problems we just need to let the value in such a way that we will get the general form $\int {{e^x}dx} = {e^x} + c$.

Note: In order to integrate any equation student must try to convert it into the general formula after the suitable substitution. Moreover, the formula must be kept in mind all the general terms and trigonometric function so as to integrate any term.