Answer is: option2
\( \quad 2x^2 + 8x + 16 \ln|x - 2| + C \)Solution:
Using long division method:
we have:
\[ \frac{4x^2}{x - 2} = 4x + 8 + \frac{16}{x - 2} \]
Integrate Term by Term
\[ \int \left( 4x + 8 + \frac{16}{x - 2} \right) dx = \int 4x \, dx + \int 8 \, dx + \int \frac{16}{x - 2} \, dx \]
\[ \int \frac{4x^2}{x - 2} \, dx = 2x^2 + 8x + 16 \ln|x - 2| + C \]
Hence option B