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

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

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

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

\frac{4}{5} arcsin (5 x) + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{1 - 25 x ^{2}} d x

Apply Trigonometric Substitution

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

= 4 \cdot \int \frac{1}{5} d u

Take the constant out:

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

= 4 \cdot \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsin (5 x)

= 4 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (5 x)

Simplify

4 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (5 x) : \frac{4}{5} arcsin (5 x)

= \frac{4}{5} arcsin (5 x)

Add a constant to the solution

= \frac{4}{5} arcsin (5 x) + C

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