Question #6c613

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Question #6c613

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Answer 1 · 2017-01-02T23:53:40

$\frac{{arctan}^{2} (2 x)}{4} + C$ $arctan^2(2x)/4+C$

Explanation:

Note I'll be using the format $arctan (x)$ $arctan(x)$ instead of $arctg (x)$ $"arctg"(x)$ .

$\int \frac{arctan (2 x)}{1 + 4 x^{2}} d x$ $intarctan(2x)/(1+4x^2)dx$

Note that since $\frac{d}{d x} arctan (x) = \frac{1}{1 + x^{2}}$ $d/dxarctan(x)=1/(1+x^2)$ , we see that $\frac{d}{d x} arctan (2 x) = \frac{1}{1 + {(2 x)}^{2}} \frac{d}{d x} (2 x) = \frac{2}{1 + 4 x^{2}}$ $d/dxarctan(2x)=1/(1+(2x)^2)d/dx(2x)=2/(1+4x^2)$ .

So, if we try the substitution $u = arctan (2 x)$ $u=arctan(2x)$ , then $d u = \frac{2}{1 + 4 x^{2}} d x$ $du=2/(1+4x^2)dx$ . We currently have $\frac{1}{1 + 4 x^{2}} d x$ $1/(1+4x^2)dx$ in the integrand, so we are just off by a factor of $2$ $2$ .

$\int \frac{arctan (2 x)}{1 + 4 x^{2}} d x = \frac{1}{2} \int arctan (2 x) \frac{2}{1 + 4 x^{2}} d x = \frac{1}{2} \int u . d u$ $intarctan(2x)/(1+4x^2)dx=1/2intarctan(2x)2/(1+4x^2)dx=1/2intucolor(white).du$

Now use the rule $\int u^{n} d u = \frac{u^{n + 1}}{n + 1}$ $intu^ndu=u^(n+1)/(n+1)$ :

$\int \frac{arctan (2 x)}{1 + 4 x^{2}} d x = \frac{1}{2} \frac{u^{2}}{2} = \frac{u^{2}}{4} = \frac{{arctan}^{2} (2 x)}{4} + C$ $intarctan(2x)/(1+4x^2)dx=1/2u^2/2=u^2/4=arctan^2(2x)/4+C$

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Question #6c613

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