$$ int \frac { 3 x ^ { 5 } + 2 x ^ { 3 } - x ^ { 2 } } { x ^ { 2 } } d x = $$\na $$ 1 8 x ^ { 6 } + 8 x ^ {…

$$ int \frac { 3 x ^ { 5 } + 2 x ^ { 3 } - x ^ { 2 } } { x ^ { 2 } } d x = $$\na $$ 1 8 x ^ { 6 } + 8 x ^ { 2 } - 2 x + c $$\nb $$ \frac { 3 } { 4 } x ^ { 4 } + x ^ { 2 } - x + c $$\nc $$ \frac { 1 5 x ^ { 4 } + 6 x ^ { 2 } - 2 x } { 2 x } + c $$\nd $$ \frac { x ^ { 6 } + x ^ { 4 } - x ^ { 3 } } { 6 x ^ { 3 } } + c $$\ne $$ 3 x ^ { 4 } + 2 x ^ { 2 } - x + c $$

$$ int \frac { 3 x ^ { 5 } + 2 x ^ { 3 } - x ^ { 2 } } { x ^ { 2 } } d x = $$\na $$ 1 8 x ^ { 6 } + 8 x ^ { 2 } - 2 x + c $$\nb $$ \frac { 3 } { 4 } x ^ { 4 } + x ^ { 2 } - x + c $$\nc $$ \frac { 1 5 x ^ { 4 } + 6 x ^ { 2 } - 2 x } { 2 x } + c $$\nd $$ \frac { x ^ { 6 } + x ^ { 4 } - x ^ { 3 } } { 6 x ^ { 3 } } + c $$\ne $$ 3 x ^ { 4 } + 2 x ^ { 2 } - x + c $$

Answer

Explanation:

Step1: Simplify the integrand

Divide each term in the numerator by (x^{2}): [ \begin{align*} \frac{3x^{5}+2x^{3}-x^{2}}{x^{2}}&=\frac{3x^{5}}{x^{2}}+\frac{2x^{3}}{x^{2}}-\frac{x^{2}}{x^{2}}\ & = 3x^{3}+2x - 1 \end{align*} ]

Step2: Integrate term - by - term

Use the power rule for integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)):

  • For (\int3x^{3}dx), we have (3\times\frac{x^{3 + 1}}{3+1}=\frac{3}{4}x^{4}) (since (\int ax^{n}dx=a\int x^{n}dx) and (\int x^{n}dx=\frac{x^{n+1}}{n + 1}+C)).
  • For (\int2xdx), we have (2\times\frac{x^{2}}{2}=x^{2}).
  • For (\int(-1)dx), we have (-x).

Combining these results, (\int(3x^{3}+2x - 1)dx=\frac{3}{4}x^{4}+x^{2}-x + C)

Answer:

B. (\frac{3}{4}x^{4}+x^{2}-x + c)