What is $\int \frac{2 x}{x^{2} + 6 x + 13} d x$ $int (2x)/(x^2+6x+13) dx$ ?

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What is $\int \frac{2 x}{x^{2} + 6 x + 13} d x$ $int (2x)/(x^2+6x+13) dx$ ?

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Answer 1 · 2016-04-28T21:43:53

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

Explanation:

A slightly different approach:

Since the derivative of the denominator is $2 x + 6$ $2x+6$ , make the following modification to the numerator:

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

Split the fraction:

$= \int \frac{2 x + 6}{x^{2} + 6 x + 13} d x - \int \frac{6}{x^{2} + 6 x + 13} d x$ $=int(2x+6)/(x^2+6x+13)dx-int6/(x^2+6x+13)dx$

For the first integral, let $u = x^{2} + 6 x + 13$ $u=x^2+6x+13$ and $d u = (2 x + 6) d x$ $du=(2x+6)dx$ .

This gives us

$\int \frac{1}{u} d u - 6 \int \frac{1}{x^{2} + 6 x + 13} d x$ $int1/udu-6int1/(x^2+6x+13)dx$

$= ln | u | - 6 \int \frac{1}{x^{2} + 6 x + 13} d x$ $=lnabsu-6int1/(x^2+6x+13)dx$

$= ln (x^{2} + 6 x + 13) - 6 \int \frac{1}{x^{2} + 6 x + 13} d x$ $=ln(x^2+6x+13)-6int1/(x^2+6x+13)dx$

Note that the absolute value signs are no longer necessary since $x^{2} + 6 x + 13 > 0$ $x^2+6x+13>0$ for all values of $x$ $x$ .

For the next integral, complete the square in the denominator.

$ln (x^{2} + 6 x + 13) - 6 \int \frac{1}{{(x + 3)}^{2} + 4} d x$ $ln(x^2+6x+13)-6int1/((x+3)^2+4)dx$

We will want to make this resemble the arctangent integral:

$\int \frac{1}{u^{2} + 1} d u = arctan (u) + C$ $int1/(u^2+1)du=arctan(u)+C$

Focusing on just $\int \frac{1}{{(x + 3)}^{2} + 4} d x$ $int1/((x+3)^2+4)dx$ , we divide everything by $4$ $4$ .

$= \int \frac{\frac{1}{4}}{\frac{{(x + 3)}^{2}}{4} + \frac{4}{4}} d x = \frac{1}{4} \int \frac{1}{\frac{{(x + 3)}^{2}}{4} + 1} d x$ $=int(1/4)/((x+3)^2/4+4/4)dx=1/4int1/((x+3)^2/4+1)dx$

To make this resemble $\int \frac{1}{u^{2} + 1} d u$ $int1/(u^2+1)du$ , we set $u = \frac{x + 3}{2}$ $u=(x+3)/2$ .

$= \frac{1}{4} \int \frac{1}{u^{2} + 1} d x$ $=1/4int1/(u^2+1)dx$

To achieve a $d u$ $du$ value, first note that $d u = \frac{1}{2} d x$ $du=1/2dx$ . Multiply the interior of the integral by $\frac{1}{2}$ $1/2$ and the exterior by $2$ $2$ .

$= \frac{1}{2} \int \frac{\frac{1}{2}}{u^{2} + 1} d x = \frac{1}{2} \int \frac{1}{u^{2} + 1} d u = \frac{1}{2} arctan (u) + C$ $=1/2int(1/2)/(u^2+1)dx=1/2int1/(u^2+1)du=1/2arctan(u)+C$

Pulling this together,

$\int \frac{1}{{(x + 3)}^{2} + 4} d x = \frac{1}{2} arctan (\frac{x + 3}{2}) + C$ $int1/((x+3)^2+4)dx=1/2arctan((x+3)/2)+C$

Combining this with the expression we came up with earlier, we see that the original integral equals

$= ln (x^{2} + 6 x + 13) - 6 \int \frac{1}{x^{2} + 6 x + 13} d x$ $=ln(x^2+6x+13)-6int1/(x^2+6x+13)dx$

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

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

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What is $\int \frac{2 x}{x^{2} + 6 x + 13} d x$ $int (2x)/(x^2+6x+13) dx$ ?

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Explanation:

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