Question

How do you integrate xe^(2x)dx?

Answer 1

You integrate it by parts:


$\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx\qquad\quad(i)}\\\\\\ \begin{array}{lcl} \mathsf{u=x}&~\Rightarrow~&\mathsf{du=dx}\\\\ \mathsf{dv=e^{2x}\,dx}&~\Leftarrow~&\mathsf{v=\dfrac{1}{2}\,e^{2x}} \end{array} \end{array}$


$\large\begin{array}{l} \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=x\cdot \frac{1}{2}\,e^{2x}-\int\!\frac{1}{2}\,e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\int\!e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\int\!\dfrac{1}{2}\cdot 2e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\cdot \frac{1}{2}\int\!e^{2x}\cdot 2\,dx} \end{array}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\int\!e^{2x}\cdot 2\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\int\!e^w\,dw\qquad(w=2x)}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\,e^w+C}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\,e^{2x}+C} \end{array}}\qquad\checkmark \end{array}$


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$\large\textsf{I hope it helps. :-)}$


Tags: integrate indefinite integral product by parts polynomial exponential composite substitution integral calculus