Question

$$ int \frac { 3 x ^ { 5 } + 2 x ^ { 3 } - x ^ { 2 } } { x ^ { 2 } } d x = $$
a $$ 1 8 x ^ { 6 } + 8 x ^ { 2 } - 2 x + c $$
b $$ \frac { 3 } { 4 } x ^ { 4 } + x ^ { 2 } - x + c $$
c $$ \frac { 1 5 x ^ { 4 } + 6 x ^ { 2 } - 2 x } { 2 x } + c $$
d $$ \frac { x ^ { 6 } + x ^ { 4 } - x ^ { 3 } } { 6 x ^ { 3 } } + c $$
e $$ 3 x ^ { 4 } + 2 x ^ { 2 } - x + c $$

Explanation:

Step1: Simplify the integrand

Divide each term in the numerator by \(x^{2}\):

$$
LATEXBLOCK0
$$

Step2: Integrate term - by - term

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

• 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. \(\frac{3}{4}x^{4}+x^{2}-x + c\)