Question

Integrate (x^2 + 2x+4)/(sqrt(x^2-4x)

Answer 1

product rule

Answer 2

hmmm let me see if this what you wrote
\[\int \frac{x^2+2x+4}{\sqrt{x^2-4x}}dx\]

Answer 3

if so let's proceed into solving this integral
how about we do completing the square of what's under the radical can we try out that

Answer 4

oh thought you are online haha

Answer 5

anyways this is a hint:
\[\int \frac{(x+1)^2+3}{\sqrt{(x-2)^2-4}}dx\]

Answer 6

One way you rewrite the given integral:
\[\begin{align*}\int\frac{x^2+2x+4}{\sqrt{x^2-4x}}\,dx&=\int\frac{x^2-4x}{\sqrt{x^2-4x}}\,dx+3\int\frac{2x-4}{\sqrt{x^2-4x}}\,dx+16\int\frac{dx}{\sqrt{x^2-4x}}\\\\
&=\int\sqrt{x^2-4x}\,dx+3\int\frac{2x-4}{\sqrt{x^2-4x}}\,dx+16\int\frac{dx}{\sqrt{x^2-4x}}\end{align*}\]
Completing the square gives, as provided earlier, \(x^2-4x=(x-2)^2-4\). From here, you can apply a trigonometric substitution for the first and third integrals, like \(x-2=2\sec u\). Meanwhile, the second integral can be simplified with a substitution of \(t=x^2-4x\), since \(dt=(2x-4)\,dx\).

@xapproachesinfinity By the way, I'm curious why you completed the square in the numerator as well. Is there something I'm missing?

Another way you can rewrite the integral: Complete the square in the denominator as before, and also write the numerator as an expression quadratic in \((x-2)\); that is,
\[\frac{x^2+2x+4}{\sqrt{x^2-4x}}=\frac{(x-2)^2+6(x-2)+12}{\sqrt{(x-2)^2-4}}\]
Again you would implement the trig sub as before, \(x-2=2\sec u\).

Answer 7

hmm i actually like the way you did it
i just completed the square so i can have (2secu+3)^2+3
when i use the trig sub 2secu=x-2
but your way is much better i see :)