What is the integral of (1/(sqrt(4-25x^2))) ?

The integral of (1/(sqrt(4-25x^2))) is 1/5 arcsin(5/2 x)+C

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integral of (1/(sqrt(4-25x^2)))

\int (\frac{1}{4 - 25 x ^{2}}) d x

\frac{1}{5} arcsin (\frac{5}{2} x) + C

Solution steps

\int \frac{1}{4 - 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 (\frac{5}{2} x)

= \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{2} x)

Simplify

= \frac{1}{5} arcsin (\frac{5}{2} x)

Add a constant to the solution

= \frac{1}{5} arcsin (\frac{5}{2} x) + C

\frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

y^{^{'}} = \frac{6 x ^{3} + y ^{3}}{x y ^{2}}

\int \frac{( x ^{2} + 4 )}{3 x ^{3} + 4 x ^{2} - 4 x} d x

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{( x ^{2} + 3 )}

y = 2 x^{3} - x^{2} + 2 (1.3)

What is the integral of (1/(sqrt(4-25x^2))) ?
The integral of (1/(sqrt(4-25x^2))) is 1/5 arcsin(5/2 x)+C