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#### karush

##### Well-known member

- Jan 31, 2012

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- Thread starter karush
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- Jan 31, 2012

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- Mar 1, 2012

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$\displaystyle \dfrac{1}{2} \int (e^{t^2}+16) \cdot 2te^{t^2} \, dt = \dfrac{1}{2} \int u \, du$

- Jan 30, 2018

- 464

We can the integral as $\int (e^{t^2}+ 16)e^{t^2}(tdt)$ and since, with the substitution $u= t^2$, $du= 2tdt$, so that $tdt= \frac{1}{2}du$ the integral becomes $\int (e^u+ 16)e^u (\frac{1}{2}du)= \frac{1}{2}\int (e^u+ 16)e^u du$. And, looking at that and remembering that at the derivative of $e^u$ is just $e^u$ again, make another substitution, $v= e^u$ o so that $dv= e^u du$ and the integral becomes $\frac{1}{2}\int (v+ 16)dv$.

(If I were very clever I would have thought of the substitution $v= e^{t^2}$ immediately but I have to do things step by step.)

That's easy to integrate! $\frac{1}{2}\int (v+ 16) dv=\frac{1}{2}(\frac{1}{2}v^2+ 16v)+ C= \frac{1}{4}v^2+ 8v+ C$ where "C" is the "constant of integration".

But now we have to go back to the original variable, t. Since $v= e^u$, $\frac{1}{4}v^2+ 8v+ C= \frac{1}{4}(e^u)^2+ 8e^u+ C= \frac{1}{4}e^{2u}+ 8e^u+ C$. And since $u= t^2$, $\frac{1}{4}e^{2u}+ 8e^u+ C= \frac{1}{4}e^{2t^2}+ 8e^{t^2}+ C$.