Evaluate $\int x ^ { 6 } \sqrt { x ^ { 7 } + 4 } d x$ .
A) $x ^ { 7 } + 4 + C$
B) $C \left( x ^ { 7 } + 4 \right) ^ { 3 / 2 }$
C) $\frac { 2 \left( x ^ { 7 } + 4 \right) ^ { 3 / 2 } } { 21 } + C$
D) $\left( x ^ { 7 } + 4 \right) ^ { 3 / 2 } + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int x^6 \sqrt{x^7 + 4} dx$, we can use a substitution method. Let $u = x^7 + 4$, then $du = 7x^6 dx$. Therefore, $x^6 dx = \frac{1}{7} du$. Substituting these into the integral gives $\int \frac{1}{7} u^{1/2} du$, which integrates to $\frac{1}{7} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{21} (x^7 + 4)^{3/2} + C$.

Evaluate ∫x6x7+4dx\int x ^ { 6 } \sqrt { x ^ { 7 } + 4 } d x∫x6x7+4​dx

Evaluate $\int x ^ { 6 } \sqrt { x ^ { 7 } + 4 } d x$