What is the integral of (5x)/(sqrt(1-25x^4)) ?

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

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integral of (5x)/(sqrt(1-25x^4))

\int \frac{5 x}{1 - 25 x ^{4}} d x

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

Solution steps

\int \frac{5 x}{1 - 25 x ^{4}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{1 - 25 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 1 - 25 u ^{2}} d u

= 5 \cdot \int \frac{1}{2 1 - 25 u ^{2}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - 25 u ^{2}} d u

Apply Trigonometric Substitution

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{5} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{d}{d x} (e^{6 x} + 3 (ln (2))^{2} e^{2 x} - e^{4 x} ln (8) - (ln (2))^{3})

3 x

f (x) = - 2 x^{4} + 8 x^{2}

x \to 0 lim (tan (π x))

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