What is the integral of x/(sqrt(1-16x^4)) ?

The integral of x/(sqrt(1-16x^4)) is 1/8 arcsin(4x^2)+C

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integral of x/(sqrt(1-16x^4))

\int \frac{x}{1 - 16 x ^{4}} d x

\frac{1}{8} arcsin (4 x^{2}) + C

Solution steps

\int \frac{x}{1 - 16 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 1 - 16 u ^{2}} d u

= \int \frac{1}{2 1 - 16 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{1 - 16 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= \frac{1}{2} \cdot \int \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 x^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 x^{2}) : \frac{1}{8} arcsin (4 x^{2})

= \frac{1}{8} arcsin (4 x^{2})

Add a constant to the solution

= \frac{1}{8} arcsin (4 x^{2}) + C

\frac{x}{( 2 x + 2 \cdot x + 3 )}

\int_{0}^{5} \frac{1}{x ^{\frac{16}{17}}} d x

\frac{d}{d θ} (5 sin (θ))

x \to 0 lim (\frac{1}{4} (4)^{4})

\int tan (3 t) d t

What is the integral of x/(sqrt(1-16x^4)) ?
The integral of x/(sqrt(1-16x^4)) is 1/8 arcsin(4x^2)+C