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Evaluate: $$\int \frac{e^x}{e^{2x} + 3e^x + 2} dx$$

My solution: Let $u = e^x$, then $$\int \frac{u}{u^2+3u+2} du=\dots$$

and $\frac{u}{(u+1)(u+2)} = \frac{A}{u+2} + \frac{B}{u+1}$ with $A = 2, B = -1$. So, $$\dots=-2 \ln |u + 2| - \ln |u+1| + C$$ resubstitute: $= 2 \ln |e^x + 2| - \ln |e^x+1| + C$. Apparently the answer is this: $$\ln(1 + e^x) - \ln(2 + e^x) + C$$

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    $u=e^x$ implies that $du=e^xdx=udx$ so $dx=\frac1udu$ – 2017-02-10
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    So basically $\frac{u^2}{u^2+3u+2} du$ will give me the answer? – 2017-02-10
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    nvm, I'm confused – 2017-02-10
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    No, the $u$ get's simplified, not squared. But I think you got it already... – 2017-02-10

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HINT The mistake is here :

enter image description here

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    Oh! Nvm I got it. Hope u didnt see my edit. Thank you :) – 2017-02-10
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No $u$ in the numerator. You forgot that $du = e^x\,dx$.