$\int \frac{1}{\sqrt{1-25 x^{2}}} d x=\frac{1}{25} \arcsin (5 x)+C$ .

Name: $\int \frac{1}{\sqrt{1-25 x^{2}}} d x=\frac{1}{25} \arcsin (5 x)+C$ .
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Question

Quizplus · Accepted Answer

The correct integral is $\int \frac{1}{\sqrt{1-25 x^{2}}} d x=\frac{1}{5} \arcsin (5 x)+C$, because when using substitution for $u=5x$, the derivative $du=5dx$ needs to be accounted for, leading to a factor of $\frac{1}{5}$ rather than $\frac{1}{25}$.

∫11−25x2dx=125arcsin⁡(5x)+C\int \frac{1}{\sqrt{1-25 x^{2}}} d x=\frac{1}{25} \arcsin (5 x)+C∫1−25x2​1​dx=251​arcsin(5x)+C

$\int \frac{1}{\sqrt{1-25 x^{2}}} d x=\frac{1}{25} \arcsin (5 x)+C$