Proper integration procedure

Question

Is this the proper procedure for the below integral?

$$\int xe^{(x^2)}dx$$

let $u=e^x$

$du=xdx$

$$=\int u^2du$$

$$={u^3\over 3}+c$$

$$={(e^x)^3\over 3}+c$$

Wolfram alpha gives me this answer $={(e^x)^2\over 2}+c$

Answer 1

Assuming you want to integrate $x(e^2x)$, then a simple wack with the integration by parts hammer, with $u = x$ and $\mathrm{d}v = e^{2x}$ so that you get $$\begin{align}\int xe^{2x} \, \mathrm{d}x &= \frac{1}{2}xe^{2x} - \frac{1}{2}\int e^{2x} \, \mathrm{d}x \\ &= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + \mathrm{C} \\ \\ & \bbox[border: blue 1px solid, 10px]{= \frac{e^{2x}}{4}\left(2x - 1\right) + \mathrm{C}} \end{align}$$

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On the other hand, if you wanted to integrate $xe^{x^2}$, then you'd make the substitution $$u= x^2 \implies \mathrm{d}u = 2x \, \mathrm{d}x$$ which makes your integral much more tractable. Your transformed integral is then $$\int \frac{1}{2}e^u \, \mathrm{d}u = \frac{1}{2}e^{u} + \mathrm{C}$$

Back substitution yields $$\bbox[border: 1px blue solid, 10px]{\int xe^{x^2}\, \mathrm{d}x = \frac{1}{2}e^{x^2} + \mathrm{C}}$$

Answer 2

The substitution used in the question is not proper. If we use the substitution $u=e^x$, then $\mathrm{d}u=e^x\,\mathrm{d}x$, not $x\,\mathrm{d}x$.

Integration by parts with $u=x$ and $v=\frac12e^{2x}$ gives $$ \begin{align} \int x(e^x)^2\,\mathrm{d}x &=\int xe^{2x}\,\mathrm{d}x\\ &=\frac12\int x\,\mathrm{d}e^{2x}\\ &=\frac12xe^{2x}-\frac12\int e^{2x}\,\mathrm{d}x\\ &=\frac12xe^{2x}-\frac14e^{2x}+C \end{align} $$

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If, on the other hand, you were trying to integrate $xe^{x^2}$, then we can use $u=x^2$ $$ \begin{align} \int xe^{x^2}\,\mathrm{d}x &=\frac12\int e^u\,\mathrm{d}u\\ &=\frac12 e^u+C\\ &=\frac12 e^{x^2}+C \end{align} $$

Answer 3

First, you really need to pay attention on how to set the brackets. As you can see, the answers are changing completely. Your procedure - substitution - is a good choice, but you made a mistake while substituting. Your $du=x dx$ is wrong it should be $du=e^x dx$, but still, this won't help you much, you're better off with $u=x^2$, just look below.

Assuming the OP wants to calculate $$ \int x(e^x)^2dx=\int xe^{2x}dx $$ then you probably want to use partial integration - also called integration by parts, which states - the $(')$ indicates the derivative - $$ \int u v^{'}dx=uv-\int v u^{'}dx $$ now we set $u=x$ and $v^{'}=e^{2x}$ which leads to $u^{'}=1$ and $v=\frac 1 2e^{2x}$ and therefore we get $$ \int u v^{'}dx=\int xe^{2x}dx=\frac 1 2xe^{2x}-\int\frac 1 2 e^{2x}dx=\frac 1 2xe^{2x}-\frac 1 4e^{2x}+c=\frac 1 2e^{2x}(x-\frac 1 2)+c $$ EDIT if you want to calculate $\int x e^{x^2}dx$

Just for the sake of completeness, now the integral with substitution $u={x^2}, du=2xdx$ $$ \int x e^{x^2}dx=\int x e^{u}\frac 1 {2x}du=\frac 1 2e^u+c=\frac1 2e^{x^2}+c $$