MHB 7.4.32 Evaluate the integral by completing the square

[karush]

7.4.32 Evaluate the integral
$\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$
ok side work to complete the square
$x^2+4x=-13$
add 4 to both sides
$x^2+4x+4=-13+4$
simplify
$(x+2)^2+9=0$
ok now whatW|A returned ≈0.03111

 

[skeeter]

7.4.32 Evaluate the integral
$\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$
ok side work to complete the square
$x^2+4x=-13$
add 4 to both sides
$x^2+4x+4=-13+4$
simplify
$(x+2)^2+9=0$
ok now what

$u = x+2 \implies du = dx$

$$\int_0^1 \dfrac{(x+2)-2}{(x+2)^2 + 9} \, dx$$

substitute and reset limits of integration ...

$$\int_2^3 \dfrac{u}{u^2 + 9} - \dfrac{2}{u^2+9} \, du$$

looks like the result will be a log and an inverse tangent ...

 

[karush]

ok little ? about where $u=x+2$ and the limits work...

$$\Biggr|\dfrac{1}{2} \ln(u^2 + 9)-\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Biggr|_2^3$$

 

[skeeter]

ok little ? about where $u=x+2$ and the limits work...

$\Bigg[\dfrac{1}{2} \ln(u^2 + 9)-\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Bigg]_2^3$

$u=x+2$

lower limit of integration is $x=0 \implies x+2 = 0+2 = 2$, the lower limit of integration reset to a $u$ value

do the same for the upper limit of integration

$\Bigg[\dfrac{1}{2} \ln(u^2 + 9)-\dfrac{2}{3}\arctan \left(\dfrac{u}{3}\right)\Bigg]_2^3$

$\left[\dfrac{1}{2}\ln(18)-\dfrac{2}{3}\arctan(1)\right] - \left[\dfrac{1}{2}\ln(13) - \dfrac{2}{3}\arctan\left(\dfrac{2}{3}\right)\right]$

$\ln{\sqrt{\dfrac{18}{13}}} - \dfrac{\pi}{6} + \dfrac{2}{3}\arctan\left(\dfrac{2}{3} \right) \approx 0.03111$