What is the integral of 1/(sqrt(1-25x^2)) ?

The integral of 1/(sqrt(1-25x^2)) is 1/5 arcsin(5x)+C

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integral of 1/(sqrt(1-25x^2))

\int \frac{1}{1 - 25 x ^{2}} d x

\frac{1}{5} arcsin (5 x) + C

Solution steps

\int \frac{1}{1 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= \int \frac{1}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsin (5 x)

= \frac{1}{5} \cdot 1 \cdot arcsin (5 x)

Simplify

= \frac{1}{5} arcsin (5 x)

Add a constant to the solution

= \frac{1}{5} arcsin (5 x) + C

f (x) = \frac{x ^{3} + x ^{2} + 6}{x}, at x = 1

\frac{\partial}{\partial x} (3 x^{4} y^{8} + 7 x^{5} y^{6})

\int 7 (1 + tan^{2} (a)) d a

h (x) = \frac{8}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

y^{^{''}} - 14 y^{^{'}} + 85 y = 0

What is the integral of 1/(sqrt(1-25x^2)) ?
The integral of 1/(sqrt(1-25x^2)) is 1/5 arcsin(5x)+C