Find the indefinite integral. $\int \frac { x } { - 4 x ^ { 2 } + 4 } d x$
A) $\frac { 1 } { - 8 x } + C$
B) $\ln \left| - 4 x ^ { 2 } + 4 \right| + C$
C) $\frac { - 1 } { 8 } \ln \left| - 4 x ^ { 2 } + 4 \right| + C$
D) $\frac { \ln \left| - 4 x ^ { 2 } + 4 \right| } { - 4 x ^ { 2 } + 4 } + C$
E) $\ln \left| - 4 x ^ { 2 } - 4 \right| + C$

Question

Quizplus · Accepted Answer

This integral can be solved by recognizing it as a standard form of an integral that results from the derivative of a logarithmic function. Specifically, the integral $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$, where $f(x) = -4x^2 + 4$ and $f'(x) = -8x$. The given integral is $\int \frac{x}{-4x^2 + 4} dx$, which can be rewritten as $\int \frac{-1/8 \cdot -8x}{-4x^2 + 4} dx$, matching the form of the logarithmic integral with a coefficient of $-1/8$ outside the integral. Therefore, the correct answer is $\frac{-1}{8} \ln|-4x^2 + 4| + C$.

Find the Indefinite Integral ∫x−4x2+4dx\int \frac { x } { - 4 x ^ { 2 } + 4 } d x∫−4x2+4x​dx

Find the Indefinite Integral $\int \frac { x } { - 4 x ^ { 2 } + 4 } d x$