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So this one seems very easy. And it really is (I guess). But I have an issue with this one. I solved it this way: \begin{align} & \int \frac{2x}{9x^2+3}dx=\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx\right) \\[10pt] = {} & \frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-x\frac{\arctan(\sqrt3x)}{\sqrt{3}}+\frac{\arctan(\sqrt3x)}{\sqrt3}\right)=\frac{\arctan(\sqrt3x)}{\sqrt3}+C \end{align} But at the same time, the textbook says, that it is equal to this: $$ \frac{1}{9}\log(3x^2+1)+C $$ Did I do something wrong? It would be quite hard for me to believ that these to functions are the same. Thanks for the responses!

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  • $\begingroup$ Try u-substitution instead. $\endgroup$ – Cameron Williams Jan 23 '16 at 19:13
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Much simpler to just set $u=9x^2+3$. So $du=18x\,dx$ and the integral becomes $$\frac19\int\frac{du}u.$$

You want to avoid trigonometric substitution if you can do $u$-substitution instead.

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Hint

$$\int \frac{2x}{9x^2+3}=\frac{1}{9}\int \frac{18x}{9x^2+3}dx.$$

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I'd have written $u = 9x^2+ 3$ so $du = 18x\,dx$ and $2x\,dx = \dfrac{du} 9$. Then you have $$ \int \frac{2x}{9x^2+3} \, dx = \frac 1 9 \int \frac{du} u = \frac 1 9 \log|u|+C = \frac 1 9 \log ( 9x^2+3) + C. $$ This is the same as $$ \frac 1 9 \log ( 3( 3x^2+1)) + C = \frac 1 9 \left( \log(3x^2+1) + \log 3 \right) + C = \frac 1 9 \log(3x^2+1) + (\text{a different } C). $$

Your initial integration by parts says $$ \frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx\right). $$ That's alright as far as it goes. I haven't attempted to figure out just what you're doing after that, but let's recall how to antidifferentiate the arctangent function: \begin{align} & \int \arctan x\,dx = \overbrace{\int u\,dx = xu - \int x\,du}^\text{integration by parts} \\[10pt] = {} & x\arctan x - \int \frac{x\,dx}{1+x^2} = x\arctan x - \int \frac{dw/2}{w} \\[10pt] = {} & x \arctan x - \frac 1 2 \log|w|+C \\[10pt] = {} & x \arctan x - \frac 1 2 \log(1+x^2) + C. \end{align} So I suspect the first term in the result just above this line (suitably adapted with your $\sqrt3$, etc.) will cancel out the first term in your integration by parts and then the result will ultimately be the same.

Generally I wouldn't integrate by parts when a simple substitution will work, unless there were some special reason to do that.

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$$\int \frac{2x}{9x^2+3}dx$$ $$2\int \frac{x}{9x^2+3}dx$$ Apply Integral Substitution $u=9x^2+3\quad \:du=18xdx$ $$2\int \frac{1}{18u}du = 2\frac{1}{18}\int \frac{1}{u}du=2\frac{1}{18}\ln \left|u\right|=2\frac{1}{18}\ln \left|9x^2+3\right|$$ So $$\int \frac{2x}{9x^2+3}dx=\color{red}{\frac{1}{9}\ln \left|9x^2+3\right|+C}$$

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