Question #b261d

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Answer 1 · 2016-12-06T19:10:47

Here are two descriptions of calculating the integral.

Finding the $r$ and $k$ requested

The expression $r[((x+k)/sqrtr)^2 +1]$ can be expanded to

$r((x+k)^2/r) + r = (x+k)^2+r$

So another way of describing what we've been asked to do is to complete the square in the denominator.

We want

$x^2+8x+20 = x^2+2kx+k^2+r$ .

So $k = 4$ and $r = 4$ .

Now we have

$int 1/(x^2+8x+20) dx = int 1/(4(((x+4)/2)^2+1)) dx$

$= 1/4 int 1/(((x+4)/2)^2+1) dx$ .

We can integrate by using the subsitution $u = (x+4)/2$ to get

$= 1/2 int 1/(u^2+1) du = 1/2 tan^-1u +C$

and finish with

$1/2 tan^-1 ((x+4)/2) +C$

OR

The method and details above are new to me. Here's how I would write the solution.

To find $int 1/(x^2+8x+20) dx$ we hope that we can use a substitution to turn the quadratic into $u^2+1$ .

To attempt to do this, complete the square.

$x^2+8x + color(white)"ssssssss" +20$

$1/2 * 8 = 4$ $" and "$ $4^2 = 16$ , so add and subtract 16 in the space above.

$x^2+8x +16 -16 + 20$ . Now simplify and factor

$(x+4)^2+4$

To make this $u^2+1$ we need to factor out the $4$ .

$4((x+4)^2/4 + 1)$

$int 1/(x^2+8x+20) dx = 1/4 int 1/(((x+4)^2/4 + 1)) dx$

With the same choice of $u$ as above, we finish the same way as above.

Perhaps the differences are subtle, but I learned the second reasoning and didn't recognize what was being requested in $r$ and $k$ .