How do you find the integral of $\int \frac{2 x}{x^{2} + 6 x + 13} d x$ $int(2x)/(x^2+6x+13) dx$

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How do you find the integral of $\int \frac{2 x}{x^{2} + 6 x + 13} d x$ $int(2x)/(x^2+6x+13) dx$

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Answer 1 · 2015-02-16T15:33:27

The answer is: $ln (x^{2} + 6 x + 13) - 3 arctan (\frac{1}{2} (x + 3)) + c$ $ln(x^2+6x+13)-3arctan(1/2(x+3))+c$

Follow my passages:

$\int \frac{2 x}{x^{2} + 6 x + 13} d x = \int \frac{2 x + 6 - 6}{x^{2} + 6 x + 13} d x =$ $int(2x)/(x^2+6x+13)dx=int(2x+6-6)/(x^2+6x+13)dx=$

$= \int \frac{2 x + 6}{x^{2} + 6 x + 13} d x - 6 \int \frac{d x}{x^{2} + 6 x + 13} = (1)$ $=int(2x+6)/(x^2+6x+13)dx-6intdx/(x^2+6x+13)=(1)$

$x^{2} + 6 x + 13 = x^{2} + 6 x + 9 + 4 = {(x + 3)}^{2} + 4 =$ $x^2+6x+13=x^2+6x+9+4=(x+3)^2+4=$

$= 4 [\frac{1}{4} {(x + 3)}^{2} + 1] = 4 [{(\frac{1}{2} (x + 3))}^{2} + 1]$ $=4[1/4(x+3)^2+1]=4[(1/2(x+3))^2+1]$ .

So:

$(1) = ln (x^{2} + 6 x + 13) - 6 \int \frac{d x}{4 [{(\frac{1}{2} (x + 3))}^{2} + 1]} =$ $(1)=ln(x^2+6x+13)-6intdx/(4[(1/2(x+3))^2+1])=$

$= ln (x^{2} + 6 x + 13) - \frac{6}{4} \int \frac{d x}{{(\frac{1}{2} (x + 3))}^{2} + 1} =$ $=ln(x^2+6x+13)-6/4intdx/((1/2(x+3))^2+1)=$

$= ln (x^{2} + 6 x + 13) - \frac{3}{2} \cdot 2 \int \frac{\frac{1}{2}}{{(\frac{1}{2} (x + 3))}^{2} + 1} d x =$ $=ln(x^2+6x+13)-3/2*2int(1/2)/((1/2(x+3))^2+1)dx=$

$= ln (x^{2} + 6 x + 13) - 3 arctan (\frac{1}{2} (x + 3)) + c$ $=ln(x^2+6x+13)-3arctan(1/2(x+3))+c$

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How do you find the integral of $\int \frac{2 x}{x^{2} + 6 x + 13} d x$ $int(2x)/(x^2+6x+13) dx$

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