What is the integral of (3x)/(sqrt(9-25x^4)) ?

The integral of (3x)/(sqrt(9-25x^4)) is 3/10 arcsin(5/3 x^2)+C

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integral of (3x)/(sqrt(9-25x^4))

\int \frac{3 x}{9 - 25 x ^{4}} d x

\frac{3}{10} arcsin (\frac{5}{3} x^{2}) + C

Solution steps

\int \frac{3 x}{9 - 25 x ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{9 - 25 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 9 - 25 u ^{2}} d u

= 3 \cdot \int \frac{1}{2 9 - 25 u ^{2}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{9 - 25 u ^{2}} d u

Apply Trigonometric Substitution

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

= 3 \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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot v

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{3} x^{2})

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{3} x^{2}) : \frac{3}{10} arcsin (\frac{5}{3} x^{2})

= \frac{3}{10} arcsin (\frac{5}{3} x^{2})

Add a constant to the solution

= \frac{3}{10} arcsin (\frac{5}{3} x^{2}) + C

f (x) = (x^{2} + 3)^{3} (x^{3} + 3)^{2}

\frac{d}{d x} (5 x^{2} cos (7 x))

f^{^{''}} (x) = x^{3} - 4 x + 1, f^{^{'}} (0) = 2, f (1) = - 2

x \to 2 lim (x^{2} - 5 x + 6 x^{2} - 4)

x \to 7 lim (\frac{x + 7}{x - 7})

What is the integral of (3x)/(sqrt(9-25x^4)) ?
The integral of (3x)/(sqrt(9-25x^4)) is 3/10 arcsin(5/3 x^2)+C