What is the integral of (16x^3)/(sqrt(1-16x^8)) ?

The integral of (16x^3)/(sqrt(1-16x^8)) is arcsin(4x^4)+C

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integral of (16x^3)/(sqrt(1-16x^8))

\int \frac{16 x ^{3}}{1 - 16 x ^{8}} d x

arcsin (4 x^{4}) + C

Solution steps

\int \frac{16 x ^{3}}{1 - 16 x ^{8}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{3}}{1 - 16 x ^{8}} d x

Apply u-substitution

: \int \frac{1}{4 1 - 16 u ^{2}} d u

= 16 \cdot \int \frac{1}{4 1 - 16 u ^{2}} d u

Take the constant out:

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

= 16 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - 16 u ^{2}} d u

Apply Trigonometric Substitution

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

= 16 \cdot \frac{1}{4} \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

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 x^{4})

Simplify

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

= arcsin (4 x^{4})

Add a constant to the solution

= arcsin (4 x^{4}) + C

x \to - 6 lim (\frac{1}{( x + 6 ) ^{4}})

y^{^{''}} - 5 y^{^{'}} + 4 y = 5 sin (5 t)

\int_{1}^{e^{x}} ln (t) d t

\int F (x) d x

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

What is the integral of (16x^3)/(sqrt(1-16x^8)) ?
The integral of (16x^3)/(sqrt(1-16x^8)) is arcsin(4x^4)+C