Integral substitusi $$ \int\frac{2x-1}{\left(x^2-2x+3\right)^5}\differentialD x $$

To solve the integral 

\[
\int \frac{2x - 1}{(x^2 - 2x + 3)^5} \, dx,
\]

we can use substitution. Let's set:

\[
u = x^2 - 2x + 3.
\]

Next, we compute the derivative of $ u $:

\[
\frac{du}{dx} = 2x - 2 \implies du = (2x - 2) \, dx.
\]

From this, we can express $ dx $:

\[
dx = \frac{du}{2x - 2} = \frac{du}{2(x - 1)}.
\]

Now, we also need to express $ 2x - 1 $ in terms of $ u $:

\[
2x - 1 = 2(x - 1) + 1.
\]

Now substituting back into the integral gives us:

\[
\int \frac{2(x - 1) + 1}{u^5} \cdot \frac{du}{2(x - 1)} = \int \left(\frac{1}{u^5} + \frac{1}{2u^5}\right) du.
\]

Now, we split the integral:

\[
\int \frac{1}{u^5} \, du + \frac{1}{2} \int \frac{1}{u^5} \, du.
\]

The integral of $ \frac{1}{u^5} $ is:

\[
\int u^{-5} \, du = -\frac{1}{4u^4} + C.
\]

Thus, we have:

\[
\int \frac{2x - 1}{(x^2 - 2x + 3)^5} \, dx = -\frac{1}{4u^4} + \frac{1}{2} \left(-\frac{1}{4u^4}\right) + C = -\frac{3}{8u^4} + C.
\]

Finally, substituting back $ u = x^2 - 2x + 3 $:

\[
= -\frac{3}{8(x^2 - 2x + 3)^4} + C.
\]

### Final Result:
\[
\int \frac{2x - 1}{(x^2 - 2x + 3)^5} \, dx = -\frac{3}{8(x^2 - 2x + 3)^4} + C.
\]

Would you like any further details or explanations? Here are some related questions to consider:

1. How do you perform substitution for integrals generally?
2. What are some properties of definite integrals?
3. Can you explain how to handle integrals with limits?
4. What is the importance of recognizing patterns in integrals?
5. How does the power rule apply to integrals? 

**Tip:** Always simplify expressions where possible before integrating to make calculations easier!

Discover how to solve the integral $$ \int\frac{2x-1}{\left(x^2-2x+3\right)^5}\differentialD x $$ using substitution. This solution includes a step-by-step explanation, relevant formulas, and key concepts in calculus. Suitable for advanced high school students and those studying calculus.

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