$\text { Use partial fractions to find } \int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x$
A)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \ln \left| x ^ { 6 } + 3 x ^ { 5 } \right| + C$
B)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \frac { 1 } { x } + \left( x ^ { 6 } + 3 x ^ { 5 } \right) + C$
C)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \frac { 1 } { x } + \ln \left| x ^ { 7 } + x ^ { 5 } \right| + C$
D)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \ln \left| x ^ { 6 } + 5 x ^ { 4 } \right| + C$
E)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \frac { 1 } { x } + \ln \left| x ^ { 6 } + 3 x ^ { 5 } \right| + C$

Question

$\text { Use partial fractions to find } \int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x$ 
A)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \ln \left| x ^ { 6 } + 3 x ^ { 5 } \right| + C$
B)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \frac { 1 } { x } + \left( x ^ { 6 } + 3 x ^ { 5 } \right) + C$
C)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \frac { 1 } { x } + \ln \left| x ^ { 7 } + x ^ { 5 } \right| + C$ 
D)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \ln \left| x ^ { 6 } + 5 x ^ { 4 } \right| + C$
E)$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 2 } ( x + 3 ) } d x = \frac { 1 } { x } + \ln \left| x ^ { 6 } + 3 x ^ { 5 } \right| + C$

Quizplus · Accepted Answer

First, we need to decompose the fraction into partial fractions:
$$\frac{6x^2+14x-3}{x^2(x+3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}$$
Multiplying both sides by $x^2(x+3)$ and simplifying, we get:
$$6x^2+14x-3=A(x+3)+B(x^2+3x)+Cx^2$$
Setting $x=0$, we get $-3=3A$ or $A=-1$. Setting $x=-3$, we get $-54=-3B$ or $B=18$. Substituting these values and equating coefficients of corresponding powers of $x$, we get $C=1$. Therefore:
$$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x = \int \left(\frac{-1}{x}+\frac{18}{x^2}+\frac{1}{x+3}\right)dx$$
Integrating each term separately, we get:
$$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x = -\ln|x|-\frac{18}{x}+\ln|x+3|+C$$
Simplifying, we get:
$$\int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x = \frac { 1 } { x } + \ln \left| \frac{x+3}{x}\right| + C$$
Which matches the answer given in choice E.

Use partial fractions to find ∫6x2+14x−3x3+3x2dx\text { Use partial fractions to find } \int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x Use partial fractions to find ∫x3+3x26x2+14x−3​dx

$\text { Use partial fractions to find } \int \frac { 6 x ^ { 2 } + 14 x - 3 } { x ^ { 3 } + 3 x ^ { 2 } } d x$