Find the integral using the indicated substitution.
$$\int ( 4 x + 3 ) ^ { 6 } d x , \quad u = 4 x + 3$$
A) $\frac { 1 } { 28 } ( 4 x + 3 ) ^ { 7 } + C$
B) $6 ( 4 x + 3 ) ^ { 7 } + C$
C) $\frac { 1 } { 6 } ( 4 x + 3 ) ^ { 7 } + C$
D) $\frac { 1 } { 7 } ( 4 x + 3 ) ^ { 7 } + C$

Question

Quizplus · Accepted Answer

Using the substitution $u = 4x + 3$, we have $du = 4 dx$. Substituting into the integral, we get:$$\int ( 4 x + 3 ) ^ { 6 } d x = \int u ^ { 6 } \frac { 1 } { 4 } du = \frac { 1 } { 4 } \cdot \frac { 1 } { 7 } u ^ { 7 } + C = \frac { 1 } { 28 } u ^ { 7 } + C = \frac { 1 } { 28 } ( 4 x + 3 ) ^ { 7 } + C$$

Find the Integral Using the Indicated Substitution ∫(4x+3)6dx,u=4x+3\int ( 4 x + 3 ) ^ { 6 } d x , \quad u = 4 x + 3∫(4x+3)6dx,u=4x+3

Find the Integral Using the Indicated Substitution $\int ( 4 x + 3 ) ^ { 6 } d x , \quad u = 4 x + 3$