Evaluate $\int x ^ { 6 } \sqrt { x ^ { 7 } + 8 } d x$ .
A) $\frac { 2 \left( x ^ { 7 } + 8 \right) ^ { 3 / 2 } } { 21 } + C$
B) $\left( x ^ { 7 } + 8 \right) ^ { 3 / 2 } + C$
C) $C \left( x ^ { 7 } + 8 \right) ^ { 3 / 2 }$
D) $x ^ { 7 } + 8 + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int x^6 \sqrt{x^7 + 8} dx$, let $u = x^7 + 8$, then $du = 7x^6 dx$. Adjusting for the $7$ in $du$, the integral becomes $\frac{1}{7} \int \sqrt{u} du$, which evaluates to $\frac{2}{7} \cdot \frac{u^{3/2}}{3/2} + C$, simplifying to $\frac{2(u^{3/2})}{21} + C$, and substituting back $u = x^7 + 8$ gives $\frac{2(x^7 + 8)^{3/2}}{21} + C$.

Evaluate ∫x6x7+8dx\int x ^ { 6 } \sqrt { x ^ { 7 } + 8 } d x∫x6x7+8​dx

Evaluate $\int x ^ { 6 } \sqrt { x ^ { 7 } + 8 } d x$