$\int 2 x \sqrt { x ^ { 2 } - 2 } d x = \frac { 2 \left( x ^ { 2 } - 2 \right) ^ { 3 / 2 } } { 3 } + C$

Name: $\int 2 x \sqrt { x ^ { 2 } - 2 } d x = \frac { 2 \left( x ^ { 2 } - 2 \right) ^ { 3 / 2 } } { 3 } + C$
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Question

Quizplus · Accepted Answer

This integral can be solved by a substitution method where $u = x^2 - 2$, making the integral transform into $\frac{2}{3}u^{3/2}$ after substitution and integration, which simplifies back to $\frac{2(x^2 - 2)^{3/2}}{3} + C$ when reverting to the original variable.

∫2xx2−2dx=2(x2−2)3/23+C\int 2 x \sqrt { x ^ { 2 } - 2 } d x = \frac { 2 \left( x ^ { 2 } - 2 \right) ^ { 3 / 2 } } { 3 } + C∫2xx2−2​dx=32(x2−2)3/2​+C

$\int 2 x \sqrt { x ^ { 2 } - 2 } d x = \frac { 2 \left( x ^ { 2 } - 2 \right) ^ { 3 / 2 } } { 3 } + C$