Find the indefinite integral. $\int \frac { 7 x ^ { 2 } } { 9 - 8 x ^ { 3 } } d x$
A) $\frac { 24 } { 7 } \ln \left| 9 - 8 x ^ { 3 } \right| + C$
B) $\ln \left| 9 - 8 x ^ { 3 } \right| + C$
C) $7 \ln \left| 9 - 8 x ^ { 3 } \right| + C$
D) $- \frac { 7 } { 24 } \ln \left| 9 - 8 x ^ { 3 } \right| + C$
E) $- \frac { 1 } { 24 } \ln \left| 9 - 8 x ^ { 3 } \right| + C$

Question

Quizplus · Accepted Answer

To solve the integral, use substitution. Let $u = 9 - 8x^3$, then $du = -24x^2 dx$. The integral becomes $\int \frac{-7}{24} \frac{1}{u} du$, which integrates to $-\frac{7}{24} \ln|u| + C$, or $-\frac{7}{24} \ln|9 - 8x^3| + C$.

Find the Indefinite Integral ∫7x29−8x3dx\int \frac { 7 x ^ { 2 } } { 9 - 8 x ^ { 3 } } d x∫9−8x37x2​dx

Find the Indefinite Integral $\int \frac { 7 x ^ { 2 } } { 9 - 8 x ^ { 3 } } d x$