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I think you have to make a substitution and then use partial fractions but im not sure how.

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That's not an integral. Anyway, try $u=e^x$. – Yonatan N Jan 30 '13 at 5:05
Try the substitution $e^x = y$, then use your partial fractions skillz. – Hanmyo Jan 30 '13 at 5:09
An integral involves bounds of integration (i.e. $-\infty$ to $\infty$ or $a$ to $b$ or whatnot). – Hanmyo Jan 30 '13 at 5:11
Are you looking for an antiderivative? – Sam DeHority Jan 30 '13 at 5:11
That is definitely an integral. Antiderivatives are also called indefinite integrals ;) – N. S. Jan 30 '13 at 5:15

Yes, make the substitution $u=e^x$. Then $du=e^x\,dx$ and we end up with $$\int \frac{du}{(u-7)(u^2+1)}.$$

Express our integrand as $\dfrac{A}{u-7}+\dfrac{Bu+C}{u^2+1}$. Bring this to the common denominator $(u-7)(u^2+1)$. The numerator is then $A(u^2+1)+(u-7)(Bu+C)$.

This numerator must be identically equal to $1$. Put $u=7$. We conclude that $50A=1$, so we know $A$. The coefficient of $u^2$ must be $0$, but it is $A+B$, so now we know $B$. Finally, from the coefficient of $u$ we can find $C$.

The integration is now straightforward. To integrate $\dfrac{Bu+C}{u^2+1}$, express as $\dfrac{Bu}{u^2+1}+\dfrac{C}{u^2+1}$. To integrate the $\dfrac{Bu}{u^2+1}$ part, let $w=u^2+1$.

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thank you this was very helpful – Pie Man Jan 30 '13 at 5:30
I think this would be a finial approach for the integral. +1 – Babak S. Jan 30 '13 at 6:48

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