How do you integrate #int (2x-2)/(x^2 - 6x +10)^(1/2)dx# using partial fractions?

Question

Ezra Adams · Accepted Answer

#[2sqrt(x^2-6x+10)]+[4arcsinh(x-3)]+C#

#int (2x-2)/sqrt(x^2 - 6x +10)dx#

#int (2x-6+4)/sqrt(x^2 - 6x +10)dx#

#int (2x-6)/sqrt(x^2 - 6x +10)dx+4int1/sqrt(x^2-6x+10)dx#

For the first integral let's #u = x^2-6x+10#

#du = 2x-6 dx#

we have #int1/sqrt(u)du#

which is #[2sqrt(u)]#

complete the square for the second

#x^2-6x+10 = x^2-6x+10 + 9 - 9 = (x-3)^2+1#

#[2sqrt(x^2-6x+10)]+4int1/sqrt((x-3)^2+1)dx#

let's #t = x-3 #

#dt = dx#

#[2sqrt(x^2-6x+10)]+4int1/sqrt(t^2+1)dx#

#[2sqrt(x^2-6x+10)]+[4arcsinh(x-3)]+C#

Isaiah Allen · Answer

undefined To integrate $\int \frac{{2x - 2}}{{(x^2 - 6x + 10)^{\frac{1}{2}}}} \, dx$ using partial fractions, you can first express the integrand as $\frac{{A}}{{(x - 3 + i)(x - 3 - i)}}$. Then, solving for A, you find that $A = -\sqrt{2}i$.

After that, you can use the substitution method to integrate, where you set $u = x - 3 + i$ and $du = dx$, leading to the integral $-\sqrt{2}i \int \frac{{du}}{{u}}$. This integral is straightforward to solve, resulting in $-\sqrt{2}i \ln|u| + C$, where $C$ is the constant of integration.

Finally, substitute back $u$ and simplify to obtain the final result of $-\sqrt{2}i \ln|x - 3 + i| + C$.