$$\int \frac{e^{2x}-2}{e^{2x}+7}\,\Bbb dx$$
I was trying:
$$\int \frac{e^{2x}}{e^{2x}+7}\,\Bbb dx -2\int \frac{1}{e^{2x}+7}\,\Bbb dx$$
$$= \frac{1}{2}\ln|e^{2x}+7| -2\int \frac{1}{(e^x)^2+(\sqrt{7})^2}\,\Bbb dx$$
$$= \frac{1}{2}\ln|e^{2x}+7|+ \frac{-2}{\sqrt{7}} \arctan\left(\frac{e^x}{\sqrt{7}}\right) +c.$$
However, the book answer says:
$$\frac{9}{14}\ln \left|e^{2x}+7\right|-\frac{2x}{7}+C.$$
Where is my mistake?
On
$\textbf{Hint:}$
$$\frac{e^{2x}-2}{e^{2x}+7} = \frac{\frac{9}{7}e^{2x}-\frac{2}{7}e^{2x}-2}{e^{2x}+7} = \frac{\frac{9}{7}e^{2x}}{e^{2x}+7} - \frac{2}{7}$$
On
The trick to these is to note that $\frac{\mathrm{d}}{\mathrm{d}x}\left(\log(e^x + 1)\right) = \frac{e^x}{e^x + 1}$, so here you have
$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(\log(e^{2x} + 7)\right) = \frac{2e^{2x}}{e^{2x} + 7} $$
To use this to solve your integral, we can express your fraction as a linear combination of $\frac{2e^{2x}}{e^{2x} + 7}$ and $1$. That is,
$$ \frac{e^{2x} - 2}{e^{2x} + 7} = c_1 \cdot \frac{2e^{2x}}{e^{2x} + 7} + c_2 \cdot 1 $$
We get $2c_1 + c_2 = 1$ and $7c_2 = -2$, so $\left(c_1, c_2\right) = \left(\frac{9}{14}, -\frac{2}{7}\right)$. Substituting, we can now integrate to get
$$ \int \frac{e^{2x} - 2}{e^{2x} + 7} \, \mathrm{d}x = \frac{9}{14}\log\left(e^{2x} + 7\right) - \frac{2}{7}x + C $$
On
$$\def\arraystretch{2}\begin{array}{c|c}u=2\,x&x=\dfrac{u}{2}\\&\mathrm{d}x=\dfrac{1}{2}\,\mathrm{d}u\end{array}$$ $$\int{\dfrac{{e}^{u}-2}{2\,\left({e}^{u}+7\right)}}{\;\mathrm{d}u}$$ \begin{gathered}\;v={e}^{u}&\\\;\mathrm{d}v={e}^{u}\,\mathrm{d}u&\end{gathered} $$\dfrac{1}{2}\int{\dfrac{v-2}{v\,\left(v+7\right)}}{\;\mathrm{d}v}$$ Grouping into sum of fractions $$\dfrac{1}{2}\int{\dfrac{1}{v+7}-\dfrac{2}{v\,\left(v+7\right)}}{\;\mathrm{d}v}$$ $$\dfrac{1}{2}\left(\int{\dfrac{1}{v+7}}{\;\mathrm{d}v}-2\int{\dfrac{1}{v\,\left(v+7\right)}}{\;\mathrm{d}v}\right)=$$ $$=\dfrac{9\,\ln\left(\left|v+7\right|\right)}{14}-\dfrac{\ln\left(\left|v\right|\right)}{7}$$ Substitute back the values $$\dfrac{9\,\ln\left({e}^{u}+7\right)}{14}-\dfrac{u}{7}$$ \begin{array}{c|c}u=2\,x&x=\dfrac{u}{2}\end{array} $$\dfrac{9\,\ln\left({e}^{2\,x}+7\right)}{14}-\dfrac{2\,x}{7}$$
On
HINT
Another possible approach to compare with: \begin{align*} \int\frac{e^{2x} - 2}{e^{2x} + 7}\mathrm{d}x & = \int\frac{e^{2x} + 7}{e^{2x} + 7}\mathrm{d}x - 9\int\frac{1}{e^{2x} + 7}\mathrm{d}x \end{align*}
Based on such relation, notice as well that: \begin{align*} \int\frac{1}{e^{2x} + 7}\mathrm{d}x & = \int\frac{e^{x}}{e^{3x} + 7e^{x}}\mathrm{d}x\\\\ & = \int\frac{\mathrm{d}(e^{x})}{(e^{x})^{3} + 7e^{x}}\\\\ & = \int\frac{\mathrm{d}u}{u^{3} + 7u} \end{align*}
where you can apply the partial fraction decomposition method to calculate it.
Can you take it from here?
You don't want to split it up like you did. Instead, think of $ - 2 $ and $ 7 $ as the leading terms (which they are as $ x \to - \infty $), and do long division: $$ \array { & \; - \frac 2 7 \hfil \\ 7 + \mathrm e ^ { 2 x } \! \! \! \! \! & ) \overline { - 2 + \mathrm e ^ { 2 x } } \hfil \\ & \quad \underline { 2 + \frac 2 7 \mathrm e ^ { 2 x } } \hfil \\ & \qquad \quad \! \frac 9 7 \mathrm e ^ { 2 x } \hfil } $$ so $$ \frac { - 2 + \mathrm e ^ { 2 x } } { 7 + \mathrm e ^ { 2 x } } = - \frac 2 7 + \frac { \frac 9 7 \mathrm e ^ { 2 x } } { 7 + \mathrm e ^ { 2 x } } \text . $$ This is how you want to split it up. The first term here gives you the $ - \frac 2 7 x $, and the second term gives you the $ \frac 9 { 1 4 } \ln ( \mathrm e ^ { 2 x } + 7 ) $. (Incidentally, you don't need the absolute value around $ \mathrm e ^ { 2 x } + 7 $, since that's always positive.)