Integral: $\displaystyle \int \dfrac{2x}{1+x}dx$. I think I have to use $u$ substitution, but I'm having trouble understanding what to do. Thank you in advance!
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$\begingroup$ Even thoug the answer is (unfortunately) already given in detail, remember: whenever the degree of the numerator is greater or equal to the denominator, first do long division! $\endgroup$ – imranfat Nov 16 '14 at 4:19
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$\begingroup$ I like the answer of Advaitha better, but you can let $u=1+x$. Then $du=dx$ and $2x=2u-2$. So we end up with $\int \frac{2u-2}{u}\,du$, This simplifies to $\int \left(2-\frac{2}{u}\right)\,du$. $\endgroup$ – André Nicolas Nov 16 '14 at 4:19
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$$\dfrac{2x}{1+x} = 2 - \dfrac2{1+x}$$ Hence, $$\int\dfrac{2x}{1+x}dx = \int2dx - \int\dfrac2{1+x}dx = 2x - 2\ln(1+x) + \text{constant}$$
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3$\begingroup$ Whenever possible, leave details of the complete answer to the OP. $\endgroup$ – Timbuc Nov 16 '14 at 4:25
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2$\begingroup$ @Timbuc: Unless the problem is 1+1 type in which case by providing a solution, one must try to emphasize each and every detail. $\endgroup$ – Nick Nov 16 '14 at 6:20
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Here's a trick against long division:
$$ \begin{align} \int \frac{2x}{1 + x}\mathrm dx &= 2\int \frac{(x + 1) - 1}{(x + 1)}\mathrm dx \\ &= 2\left(\int\mathrm dx -\int\frac{\mathrm d(x+1)}{x+1}\right)\\ &= 2\Big(x - \ln|x+1|\Big) \color{grey}{+ \mathcal C} \end{align}$$
Note that $2x + \ln\left|\frac{1}{x^2 + 2x + 1}\right| + \color{grey}{\mathcal C}$ is also a valid answer =)
