What is the integral of 1/(sqrt(1-49x^2)) ?

The integral of 1/(sqrt(1-49x^2)) is 1/7 arcsin(7x)+C

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integral of 1/(sqrt(1-49x^2))

\int \frac{1}{1 - 49 x ^{2}} d x

\frac{1}{7} arcsin (7 x) + C

Solution steps

\int \frac{1}{1 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= \int \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsin (7 x)

= \frac{1}{7} \cdot 1 \cdot arcsin (7 x)

Simplify

= \frac{1}{7} arcsin (7 x)

Add a constant to the solution

= \frac{1}{7} arcsin (7 x) + C

\int \frac{x ^{2} - 25}{x - 5} d x

\frac{\partial}{\partial y} (4 x^{3} y^{3})

y = 2 x^{2} - 1, (0, 1)

x \to 4 lim (4 x^{2} + 5 x + 1)

\frac{d}{d x} (3 x + 3)

What is the integral of 1/(sqrt(1-49x^2)) ?
The integral of 1/(sqrt(1-49x^2)) is 1/7 arcsin(7x)+C