Find the indefinite integral $\int t ^ { 3 } \left( 5 + t ^ { 4 } \right) ^ { 5 } d t$
A) $\frac { \left( 5 + t ^ { 3 } \right) ^ { 6 } } { 24 } + C$
B) $\frac { \left( 5 + t ^ { 4 } \right) ^ { 6 } } { 24 } + C$
C) $\frac { \left( 5 + t ^ { 4 } \right) ^ { 6 } } { 6 } + C$
D) $24 \left( 5 + t ^ { 4 } \right) ^ { 6 } + C$
E) $\frac { \left( 5 + t ^ { 4 } \right) ^ { 6 } } { 16 } + C$

Question

Find the indefinite integral $\int t ^ { 3 } \left( 5 + t ^ { 4 } \right) ^ { 5 } d t$ 
A) $\frac { \left( 5 + t ^ { 3 } \right) ^ { 6 } } { 24 } + C$
B) $\frac { \left( 5 + t ^ { 4 } \right) ^ { 6 } } { 24 } + C$
C) $\frac { \left( 5 + t ^ { 4 } \right) ^ { 6 } } { 6 } + C$
D) $24 \left( 5 + t ^ { 4 } \right) ^ { 6 } + C$
E) $\frac { \left( 5 + t ^ { 4 } \right) ^ { 6 } } { 16 } + C$

Quizplus · Accepted Answer

We can use the substitution $u = 5+t^4$ and then $du = 4t^3 dt$.
Substituting, we get:
$$\int t^3 (5+t^4)^5 dt = \frac{1}{4} \int (5+t^4)^5 (4t^3) dt$$
$$= \frac{1}{4} \int u^5 du = \frac{1}{4} \cdot \frac{1}{6} u^6 + C$$
$$= \frac{1}{24} (5+t^4)^6 + C$$
Therefore, the answer is $\boxed{\text{(B)}}$.

Find the Indefinite Integral ∫t3(5+t4)5dt\int t ^ { 3 } \left( 5 + t ^ { 4 } \right) ^ { 5 } d t∫t3(5+t4)5dt

Find the Indefinite Integral $\int t ^ { 3 } \left( 5 + t ^ { 4 } \right) ^ { 5 } d t$