Solve the problem by integration.
-Find the first-quadrant area bounded by $y = \frac { 1 } { x ^ { 3 } + 4 x ^ { 2 } + 3 x } , x = 1$, and $x = 4$.
A) $\frac { 1 } { 6 } \ln \frac { 224 } { 125 }$
B) $6 \ln \frac { 125 } { 1792 }$
C) $\frac { 1 } { 3 } \ln \frac { 32 } { 49 }$
D) $\frac { 1 } { 6 } \ln \frac { 125 } { 1792 }$

Question

Quizplus · Accepted Answer

To find the area, we integrate the given function from $x = 1$ to $x = 4$. The integral of $y = \frac { 1 } { x ^ { 3 } + 4 x ^ { 2 } + 3 x }$ can be simplified by noticing that the denominator is a factor of $x(x^2 + 4x + 3)$, which further factors into $x(x+1)(x+3)$. Thus, the integral becomes $\int_{1}^{4} \frac{1}{x(x+1)(x+3)} dx$. Using partial fraction decomposition, this can be expressed as a sum of simpler fractions, each of which can be integrated individually. The result of this integration gives the area under the curve, which matches the expression in choice A: $\frac { 1 } { 6 } \ln \frac { 224 } { 125 }$.

Solve the Problem by Integration y=1x3+4x2+3x,x=1y = \frac { 1 } { x ^ { 3 } + 4 x ^ { 2 } + 3 x } , x = 1y=x3+4x2+3x1​,x=1

Solve the Problem by Integration $y = \frac { 1 } { x ^ { 3 } + 4 x ^ { 2 } + 3 x } , x = 1$