Determine the integral by making an appropriate substitution.
-$\int 8 x \sqrt [ 3 ] { 4 x ^ { 2 } - 9 }$ dx
A) $\frac { \left( 4 x ^ { 2 } + 9 \right) ^ { 4 } } { 16 }$ + C
B) $\frac { \left( 4 x ^ { 2 } - 9 \right) ^ { 4 / 3 } } { 3 }$ + C
C) $\frac { 3 \left( 4 x ^ { 2 } + 9 \right) ^ { 4 / 3 } } { 4 }$ + C
D) $\frac { 3 \left( 4 x ^ { 2 } - 9 \right) ^ { 4 / 3 } } { 4 }$ + C

Question

Determine the integral by making an appropriate substitution. 
-$\int 8 x \sqrt [ 3 ] { 4 x ^ { 2 } - 9 }$ dx
A) $\frac { \left( 4 x ^ { 2 } + 9 \right) ^ { 4 } } { 16 }$ + C
B) $\frac { \left( 4 x ^ { 2 } - 9 \right) ^ { 4 / 3 } } { 3 }$ + C
C) $\frac { 3 \left( 4 x ^ { 2 } + 9 \right) ^ { 4 / 3 } } { 4 }$ + C
D) $\frac { 3 \left( 4 x ^ { 2 } - 9 \right) ^ { 4 / 3 } } { 4 }$ + C

Quizplus · Accepted Answer

Let $u = 4x^2 - 9$, then $du = 8x dx$. The integral becomes $\int u^{1/3} du$, which evaluates to $\frac{3}{4} u^{4/3} + C$, or $\frac{3(4x^2 - 9)^{4/3}}{4} + C$.

Determine the Integral by Making an Appropriate Substitution ∫8x4x2−93\int 8 x \sqrt [ 3 ] { 4 x ^ { 2 } - 9 }∫8x34x2−9​

Determine the Integral by Making an Appropriate Substitution $\int 8 x \sqrt [ 3 ] { 4 x ^ { 2 } - 9 }$