
MHB Master
#1
October 7th, 2019,
16:23
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
WA returned ≈0.03111
Last edited by karush; October 7th, 2019 at 16:37.

October 7th, 2019 16:23
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MHB Craftsman
#2
October 7th, 2019,
16:45
Originally Posted by
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 what
$u = x+2 \implies du = dx$
$ \displaystyle \int_0^1 \dfrac{(x+2)2}{(x+2)^2 + 9} \, dx$
substitute and reset limits of integration ...
$ \displaystyle \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 ...

MHB Master
#3
October 7th, 2019,
17:43
Thread Author
ok little ??? about where $u=x+2$ and the limits work...
$$\Biggr\dfrac{1}{2} \ln(u^2 + 9)\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Biggr_2^3$$

MHB Craftsman
#4
October 7th, 2019,
19:28
Originally Posted by
karush
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
$\Bigg[\dfrac{1}{2} \ln(u^2 + 9)\dfrac{2}{3}\arctan \left(\dfrac{u}{3}\right)\Bigg]_2^3$
$\left[\dfrac{1}{2}\ln(18)\dfrac{2}{3}\arctan(1)\right]  \left[\dfrac{1}{2}\ln(13)  \dfrac{2}{3}\arctan\left(\dfrac{2}{3}\right)\right]$
$\ln{\sqrt{\dfrac{18}{13}}}  \dfrac{\pi}{6} + \dfrac{2}{3}\arctan\left(\dfrac{2}{3} \right) \approx 0.03111$