$\int \frac { x ^ { 3 } - 3 x ^ { 2 } + 7 } { x } d x = \frac { x ^ { 3 } } { 3 } - \frac { 3 x ^ { 2 } } { 2 } + 7 \ln | x | + C$

Question

Quizplus · Accepted Answer

The correct integral should be $\int \frac { x ^ { 3 } - 3 x ^ { 2 } + 7 } { x } d x = \frac { x ^ { 2 } } { 2 } - \frac { 3 x } { 2 } + 7 \ln | x | + C$, because when you divide each term in the numerator by $x$, you get $x^2 - 3x + 7$, and then integrate each term separately.

∫x3−3x2+7xdx=x33−3x22+7ln⁡∣x∣+C\int \frac { x ^ { 3 } - 3 x ^ { 2 } + 7 } { x } d x = \frac { x ^ { 3 } } { 3 } - \frac { 3 x ^ { 2 } } { 2 } + 7 \ln | x | + C∫xx3−3x2+7​dx=3x3​−23x2​+7ln∣x∣+C

$\int \frac { x ^ { 3 } - 3 x ^ { 2 } + 7 } { x } d x = \frac { x ^ { 3 } } { 3 } - \frac { 3 x ^ { 2 } } { 2 } + 7 \ln | x | + C$