$\text { Find the indefinite integral } \int \frac { x ^ { 2 } } { 4 x ^ { 3 } + 9 } d x \text {. }$
A) $\frac { x ^ { 3 } } { 4 x ^ { 4 } + 9 x } + C$
B) $\ln \left| 4 x ^ { 3 } + 9 \right| + C$
C) $\frac { 1 } { 4 } \ln \left| 4 x ^ { 3 } + 9 \right| + C$
D) $\frac { 1 } { 12 } \ln \left| 4 x ^ { 3 } + 9 \right| + C$
E) $12 \ln \left| 4 x ^ { 3 } + 9 \right| + C$

Question

$\text { Find the indefinite integral } \int \frac { x ^ { 2 } } { 4 x ^ { 3 } + 9 } d x \text {. }$ 
A) $\frac { x ^ { 3 } } { 4 x ^ { 4 } + 9 x } + C$
B) $\ln \left| 4 x ^ { 3 } + 9 \right| + C$
C) $\frac { 1 } { 4 } \ln \left| 4 x ^ { 3 } + 9 \right| + C$
D) $\frac { 1 } { 12 } \ln \left| 4 x ^ { 3 } + 9 \right| + C$
E) $12 \ln \left| 4 x ^ { 3 } + 9 \right| + C$

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = 4x^3 + 9 $, then $ du = 12x^2 \, dx $. Thus, $ dx = \frac{1}{12x^2} \, du $. Substituting these into the integral gives $\int \frac{x^2}{u} \cdot \frac{1}{12x^2} \, du = \frac{1}{12} \int \frac{1}{u} \, du = \frac{1}{12} \ln |u| + C = \frac{1}{12} \ln |4x^3 + 9| + C$.

Find the indefinite integral ∫x24x3+9dx. \text { Find the indefinite integral } \int \frac { x ^ { 2 } } { 4 x ^ { 3 } + 9 } d x \text {. } Find the indefinite integral ∫4x3+9x2​dx.

$\text { Find the indefinite integral } \int \frac { x ^ { 2 } } { 4 x ^ { 3 } + 9 } d x \text {. }$