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

Given $u = 4x + 3$, the derivative $du = 4dx$, which implies $dx = \frac{1}{4}du$. Substituting into the integral gives $\int u^6 \cdot \frac{1}{4}du$, which integrates to $\frac{1}{4} \cdot \frac{1}{7}u^7 + C$, or $\frac{1}{28}(4x + 3)^7 + C$.

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 $\int ( 4 x + 3 ) ^ { 6 } d x , \quad u = 4 x + 3$