How do you integrate $\int e^{x} e^{x}$ $int e^xe^x$ using substitution?

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How do you integrate $\int e^{x} e^{x}$ $int e^xe^x$ using substitution?

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Answer 1 · 2016-11-20T14:40:13

$\int e^{x} e^{x} d x$ $inte^xe^xdx$

Method 1 - Immediate Substitution

Make the substitution $u = e^{x}$ $u=e^x$ , which implies that $d u = e^{x} d x$ $du=e^xdx$ , so

$\int e^{x} e^{x} d x = \int (e^{x}) (e^{x} d x) = \int u d u = \frac{u^{2}}{2} = \frac{{(e^{x})}^{2}}{2} = \frac{e^{2 x}}{2} + C$ $inte^xe^xdx=int(e^x)(e^xdx)=intudu=u^2/2=(e^x)^2/2=e^(2x)/2+C$

Method 2 - Simplification, then Substitution

Use the rule $a^{b} (a^{c}) = a^{b + c}$ $a^b(a^c)=a^(b+c)$ to rewrite the integral as

$\int e^{x} e^{x} d x = \int e^{2 x} d x$ $inte^xe^xdx=inte^(2x)dx$

Now substitute $u = 2 x$ $u=2x$ so $d u = 2 d x$ $du=2dx$ :

$\int e^{2 x} d x = \frac{1}{2} \int (e^{2 x}) (2 d x) = \frac{1}{2} \int e^{u} d u$ $inte^(2x)dx=1/2int(e^(2x))(2dx)=1/2inte^udu$

Since $\int e^{u} d u = e^{u}$ $inte^udu=e^u$ :

$\frac{1}{2} \int e^{u} d u = \frac{1}{2} e^{u} = \frac{e^{u}}{2} = \frac{e^{2 x}}{2} + C$ $1/2inte^udu=1/2e^u=e^u/2=e^(2x)/2+C$

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How do you integrate $\int e^{x} e^{x}$ $int e^xe^x$ using substitution?

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