Evaluate the integral. $\int x \left( 5 x ^ { 2 } + 2 \right) ^ { 3 } \mathrm {~d} x$
A) $\frac { ( 5 x + 2 ) ^ { 2 } } { 5 } + C$
B) $\frac { x \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 40 } + C$
C) $\frac { \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 40 } + C$
D) $\frac { \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 20 } + C$
E) $\frac { \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 40 }$

Question

Evaluate the integral.  $\int x \left( 5 x ^ { 2 } + 2 \right) ^ { 3 } \mathrm {~d} x$ 
A)  $\frac { ( 5 x + 2 ) ^ { 2 } } { 5 } + C$
B)   $\frac { x \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 40 } + C$
C)   $\frac { \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 40 } + C$
D)  $\frac { \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 20 } + C$
E)   $\frac { \left( 5 x ^ { 2 } + 2 \right) ^ { 4 } } { 40 }$

Quizplus · Accepted Answer

To solve the integral $\int x \left( 5 x ^ { 2 } + 2 \right) ^ { 3 } \mathrm {~d} x$, we can use a substitution method. Let $u = 5x^2 + 2$, then $du = 10x dx$. The integral becomes $\frac{1}{10} \int u^3 du$, which evaluates to $\frac{1}{10} \cdot \frac{u^4}{4} + C = \frac{\left( 5 x ^ { 2 } + 2 \right) ^ { 4 }}{40} + C$.

Evaluate the Integral ∫x(5x2+2)3 dx\int x \left( 5 x ^ { 2 } + 2 \right) ^ { 3 } \mathrm {~d} x∫x(5x2+2)3 dx

Evaluate the Integral $\int x \left( 5 x ^ { 2 } + 2 \right) ^ { 3 } \mathrm {~d} x$